Attempt on modified harmonic series












2















Does the series $left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9} right)+ldots quad$ converge?




Here's my attempt at a solution: begin{eqnarray*}left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9} right)+ldots &= sum_{n=1}^{infty}frac{1}{n}-2sum_{n=1}^{infty}frac{1}{3n}\&=sum_{n=1}^{infty}frac{1}{3n} \ &=frac{1}{3}sum_{n=1}^{infty}frac{1}{n} end{eqnarray*}



As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of the requiered series.



Is this right? Which other convergence tests could be used?

Thanks in advance.






calculus sequences-and-series proof-verification divergent-series







share|cite|improve this question







New contributor




Raúl Astete is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
    – Masacroso
    48 mins ago












  • @Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
    – Saad
    42 mins ago
















2















Does the series $left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9} right)+ldots quad$ converge?




Here's my attempt at a solution: begin{eqnarray*}left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9} right)+ldots &= sum_{n=1}^{infty}frac{1}{n}-2sum_{n=1}^{infty}frac{1}{3n}\&=sum_{n=1}^{infty}frac{1}{3n} \ &=frac{1}{3}sum_{n=1}^{infty}frac{1}{n} end{eqnarray*}



As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of the requiered series.



Is this right? Which other convergence tests could be used?

Thanks in advance.






calculus sequences-and-series proof-verification divergent-series







share|cite|improve this question







New contributor




Raúl Astete is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
    – Masacroso
    48 mins ago












  • @Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
    – Saad
    42 mins ago














2












2








2








Does the series $left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9} right)+ldots quad$ converge?




Here's my attempt at a solution: begin{eqnarray*}left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9} right)+ldots &= sum_{n=1}^{infty}frac{1}{n}-2sum_{n=1}^{infty}frac{1}{3n}\&=sum_{n=1}^{infty}frac{1}{3n} \ &=frac{1}{3}sum_{n=1}^{infty}frac{1}{n} end{eqnarray*}



As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of the requiered series.



Is this right? Which other convergence tests could be used?

Thanks in advance.






calculus sequences-and-series proof-verification divergent-series







share|cite|improve this question







New contributor




Raúl Astete is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












Does the series $left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9} right)+ldots quad$ converge?




Here's my attempt at a solution: begin{eqnarray*}left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9} right)+ldots &= sum_{n=1}^{infty}frac{1}{n}-2sum_{n=1}^{infty}frac{1}{3n}\&=sum_{n=1}^{infty}frac{1}{3n} \ &=frac{1}{3}sum_{n=1}^{infty}frac{1}{n} end{eqnarray*}



As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of the requiered series.



Is this right? Which other convergence tests could be used?

Thanks in advance.






calculus sequences-and-series proof-verification divergent-series




calculus sequences-and-series proof-verification divergent-series






share|cite|improve this question







New contributor




Raúl Astete is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Raúl Astete is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




Raúl Astete is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 1 hour ago









Raúl Astete

256




256




New contributor




Raúl Astete is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Raúl Astete is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Raúl Astete is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
    – Masacroso
    48 mins ago












  • @Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
    – Saad
    42 mins ago


















  • your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
    – Masacroso
    48 mins ago












  • @Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
    – Saad
    42 mins ago
















your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
– Masacroso
48 mins ago






your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
– Masacroso
48 mins ago














@Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
– Saad
42 mins ago




@Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
– Saad
42 mins ago










3 Answers
3






active

oldest

votes


















4














Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
$$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
Simplify what is in the parentheses and then evaluate in the usual way.



Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.






share|cite|improve this answer































    2














    The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.



    Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
    $$
    S_n = H_{3n} - frac 23 H_n,
    $$
    then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
    $$
    S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
    $$






    share|cite|improve this answer





























      1














      Yes, your answer is correct.



      By using Cauchy's Convergence test, for $sum_{n=1}^inftydfrac1n$



      Taking $$S_{2n}-S_n=sum_{n=1}^{2n}dfrac1n-sum_{n=1}^ndfrac1n=dfrac{1}{n+1}+dfrac{1}{n+2}+....+dfrac{1}{2n}gedfrac{1}{2n}+....+dfrac{1}{2n}=dfrac{n}{2n}=dfrac{1}{2}$$



      Since, there is no number $N$ that satisfies Cauchy's condition, the series diverges.






      share|cite|improve this answer





















        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });






        Raúl Astete is a new contributor. Be nice, and check out our Code of Conduct.










        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063422%2fattempt-on-modified-harmonic-series%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4














        Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
        $$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
        Simplify what is in the parentheses and then evaluate in the usual way.



        Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.






        share|cite|improve this answer




























          4














          Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
          $$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
          Simplify what is in the parentheses and then evaluate in the usual way.



          Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.






          share|cite|improve this answer


























            4












            4








            4






            Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
            $$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
            Simplify what is in the parentheses and then evaluate in the usual way.



            Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.






            share|cite|improve this answer














            Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
            $$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
            Simplify what is in the parentheses and then evaluate in the usual way.



            Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 54 mins ago

























            answered 1 hour ago









            Ben W

            1,825514




            1,825514























                2














                The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.



                Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
                $$
                S_n = H_{3n} - frac 23 H_n,
                $$
                then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
                $$
                S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
                $$






                share|cite|improve this answer


























                  2














                  The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.



                  Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
                  $$
                  S_n = H_{3n} - frac 23 H_n,
                  $$
                  then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
                  $$
                  S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
                  $$






                  share|cite|improve this answer
























                    2












                    2








                    2






                    The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.



                    Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
                    $$
                    S_n = H_{3n} - frac 23 H_n,
                    $$
                    then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
                    $$
                    S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
                    $$






                    share|cite|improve this answer












                    The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.



                    Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
                    $$
                    S_n = H_{3n} - frac 23 H_n,
                    $$
                    then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
                    $$
                    S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
                    $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 50 mins ago









                    xbh

                    5,6951522




                    5,6951522























                        1














                        Yes, your answer is correct.



                        By using Cauchy's Convergence test, for $sum_{n=1}^inftydfrac1n$



                        Taking $$S_{2n}-S_n=sum_{n=1}^{2n}dfrac1n-sum_{n=1}^ndfrac1n=dfrac{1}{n+1}+dfrac{1}{n+2}+....+dfrac{1}{2n}gedfrac{1}{2n}+....+dfrac{1}{2n}=dfrac{n}{2n}=dfrac{1}{2}$$



                        Since, there is no number $N$ that satisfies Cauchy's condition, the series diverges.






                        share|cite|improve this answer


























                          1














                          Yes, your answer is correct.



                          By using Cauchy's Convergence test, for $sum_{n=1}^inftydfrac1n$



                          Taking $$S_{2n}-S_n=sum_{n=1}^{2n}dfrac1n-sum_{n=1}^ndfrac1n=dfrac{1}{n+1}+dfrac{1}{n+2}+....+dfrac{1}{2n}gedfrac{1}{2n}+....+dfrac{1}{2n}=dfrac{n}{2n}=dfrac{1}{2}$$



                          Since, there is no number $N$ that satisfies Cauchy's condition, the series diverges.






                          share|cite|improve this answer
























                            1












                            1








                            1






                            Yes, your answer is correct.



                            By using Cauchy's Convergence test, for $sum_{n=1}^inftydfrac1n$



                            Taking $$S_{2n}-S_n=sum_{n=1}^{2n}dfrac1n-sum_{n=1}^ndfrac1n=dfrac{1}{n+1}+dfrac{1}{n+2}+....+dfrac{1}{2n}gedfrac{1}{2n}+....+dfrac{1}{2n}=dfrac{n}{2n}=dfrac{1}{2}$$



                            Since, there is no number $N$ that satisfies Cauchy's condition, the series diverges.






                            share|cite|improve this answer












                            Yes, your answer is correct.



                            By using Cauchy's Convergence test, for $sum_{n=1}^inftydfrac1n$



                            Taking $$S_{2n}-S_n=sum_{n=1}^{2n}dfrac1n-sum_{n=1}^ndfrac1n=dfrac{1}{n+1}+dfrac{1}{n+2}+....+dfrac{1}{2n}gedfrac{1}{2n}+....+dfrac{1}{2n}=dfrac{n}{2n}=dfrac{1}{2}$$



                            Since, there is no number $N$ that satisfies Cauchy's condition, the series diverges.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 1 hour ago









                            Key Flex

                            7,56241232




                            7,56241232






















                                Raúl Astete is a new contributor. Be nice, and check out our Code of Conduct.










                                draft saved

                                draft discarded


















                                Raúl Astete is a new contributor. Be nice, and check out our Code of Conduct.













                                Raúl Astete is a new contributor. Be nice, and check out our Code of Conduct.












                                Raúl Astete is a new contributor. Be nice, and check out our Code of Conduct.
















                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.





                                Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                Please pay close attention to the following guidance:


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063422%2fattempt-on-modified-harmonic-series%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                -

                                Popular posts from this blog

                                Plaza Victoria

                                Image
                                Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                                Read more

                                Puebla de Zaragoza

                                Image
                                Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
                                Read more

                                Musa

                                Image
                                Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
                                Read more