Can we Combine two convergent infinite series as follows












1












$begingroup$


Given two infinite series $S_0 = sum_{n=1}^{infty}frac{1}{F(x, n)^{p}}$ and $S_1 = sum_{n=1}^{infty}frac{1}{F'(x, n)^{p}}$.



Let both of these series be 'convergent' to the same real value $K$ for some values of $x$.



Assuming for $x = x_0$ this is the case. Thus,



$S_0 = sum_{n=1}^{infty}frac{1}{F(x_0, n)^{p}} = K$



$S_1 = sum_{n=1}^{infty}frac{1}{F'(x_0, n)^{p}} = K$



Now, can we claim that for new series used by combining the corresponding terms of $S_0$ and $S_1$ are also convergent, i.e.:



$S_2 = sum_{n=1}^{infty}(frac{1}{F(x_0, n)^{p}}) - (frac{1}{F'(x_0, n)^{p}}) = 0$



$S_3 = sum_{n=1}^{infty}(frac{1}{F(x_0, n)^{p}}) + (frac{1}{F'(x_0, n)^{p}}) = 2K$



I assume this is true, but I am not certain.






real-analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    If $S_n to S $ and $T_n to T$ then $S_n - T_n to S -T$. Letting these denote your partial sum sequences we see its OK.
    $endgroup$
    – RRL
    Dec 14 '18 at 8:12
















1












$begingroup$


Given two infinite series $S_0 = sum_{n=1}^{infty}frac{1}{F(x, n)^{p}}$ and $S_1 = sum_{n=1}^{infty}frac{1}{F'(x, n)^{p}}$.



Let both of these series be 'convergent' to the same real value $K$ for some values of $x$.



Assuming for $x = x_0$ this is the case. Thus,



$S_0 = sum_{n=1}^{infty}frac{1}{F(x_0, n)^{p}} = K$



$S_1 = sum_{n=1}^{infty}frac{1}{F'(x_0, n)^{p}} = K$



Now, can we claim that for new series used by combining the corresponding terms of $S_0$ and $S_1$ are also convergent, i.e.:



$S_2 = sum_{n=1}^{infty}(frac{1}{F(x_0, n)^{p}}) - (frac{1}{F'(x_0, n)^{p}}) = 0$



$S_3 = sum_{n=1}^{infty}(frac{1}{F(x_0, n)^{p}}) + (frac{1}{F'(x_0, n)^{p}}) = 2K$



I assume this is true, but I am not certain.






real-analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    If $S_n to S $ and $T_n to T$ then $S_n - T_n to S -T$. Letting these denote your partial sum sequences we see its OK.
    $endgroup$
    – RRL
    Dec 14 '18 at 8:12














1












1








1





$begingroup$


Given two infinite series $S_0 = sum_{n=1}^{infty}frac{1}{F(x, n)^{p}}$ and $S_1 = sum_{n=1}^{infty}frac{1}{F'(x, n)^{p}}$.



Let both of these series be 'convergent' to the same real value $K$ for some values of $x$.



Assuming for $x = x_0$ this is the case. Thus,



$S_0 = sum_{n=1}^{infty}frac{1}{F(x_0, n)^{p}} = K$



$S_1 = sum_{n=1}^{infty}frac{1}{F'(x_0, n)^{p}} = K$



Now, can we claim that for new series used by combining the corresponding terms of $S_0$ and $S_1$ are also convergent, i.e.:



$S_2 = sum_{n=1}^{infty}(frac{1}{F(x_0, n)^{p}}) - (frac{1}{F'(x_0, n)^{p}}) = 0$



$S_3 = sum_{n=1}^{infty}(frac{1}{F(x_0, n)^{p}}) + (frac{1}{F'(x_0, n)^{p}}) = 2K$



I assume this is true, but I am not certain.






real-analysis







share|cite|improve this question









$endgroup$




Given two infinite series $S_0 = sum_{n=1}^{infty}frac{1}{F(x, n)^{p}}$ and $S_1 = sum_{n=1}^{infty}frac{1}{F'(x, n)^{p}}$.



Let both of these series be 'convergent' to the same real value $K$ for some values of $x$.



Assuming for $x = x_0$ this is the case. Thus,



$S_0 = sum_{n=1}^{infty}frac{1}{F(x_0, n)^{p}} = K$



$S_1 = sum_{n=1}^{infty}frac{1}{F'(x_0, n)^{p}} = K$



Now, can we claim that for new series used by combining the corresponding terms of $S_0$ and $S_1$ are also convergent, i.e.:



$S_2 = sum_{n=1}^{infty}(frac{1}{F(x_0, n)^{p}}) - (frac{1}{F'(x_0, n)^{p}}) = 0$



$S_3 = sum_{n=1}^{infty}(frac{1}{F(x_0, n)^{p}}) + (frac{1}{F'(x_0, n)^{p}}) = 2K$



I assume this is true, but I am not certain.






real-analysis




real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 14 '18 at 8:05









TheoryQuest1TheoryQuest1

1498




1498








  • 1




    $begingroup$
    If $S_n to S $ and $T_n to T$ then $S_n - T_n to S -T$. Letting these denote your partial sum sequences we see its OK.
    $endgroup$
    – RRL
    Dec 14 '18 at 8:12














  • 1




    $begingroup$
    If $S_n to S $ and $T_n to T$ then $S_n - T_n to S -T$. Letting these denote your partial sum sequences we see its OK.
    $endgroup$
    – RRL
    Dec 14 '18 at 8:12








1




1




$begingroup$
If $S_n to S $ and $T_n to T$ then $S_n - T_n to S -T$. Letting these denote your partial sum sequences we see its OK.
$endgroup$
– RRL
Dec 14 '18 at 8:12




$begingroup$
If $S_n to S $ and $T_n to T$ then $S_n - T_n to S -T$. Letting these denote your partial sum sequences we see its OK.
$endgroup$
– RRL
Dec 14 '18 at 8:12










1 Answer
1






active

oldest

votes


















2












$begingroup$

Yes, since your $x_0$ is fixed anyway we can just use the standard rule of adding two convergent series:
$$
sum_{n=1}^infty (a_n + b_n) = sum_{n=1}^infty a_n + sum_{n=1}^infty b_n.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. I assume by extension its also true for multiplying and division too? And also were $x$ is a complex number too ?
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:16






  • 2




    $begingroup$
    For division you'd need that the divisor doesn't converge to $0$. It also works for complex number. The proofs of these facts are pretty standard and shouldn't be too hard to find.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:17










  • $begingroup$
    Thanks again. I am just beginning in analysis as a self-interest. my background is CS.
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:18











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039092%2fcan-we-combine-two-convergent-infinite-series-as-follows%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Yes, since your $x_0$ is fixed anyway we can just use the standard rule of adding two convergent series:
$$
sum_{n=1}^infty (a_n + b_n) = sum_{n=1}^infty a_n + sum_{n=1}^infty b_n.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. I assume by extension its also true for multiplying and division too? And also were $x$ is a complex number too ?
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:16






  • 2




    $begingroup$
    For division you'd need that the divisor doesn't converge to $0$. It also works for complex number. The proofs of these facts are pretty standard and shouldn't be too hard to find.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:17










  • $begingroup$
    Thanks again. I am just beginning in analysis as a self-interest. my background is CS.
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:18
















2












$begingroup$

Yes, since your $x_0$ is fixed anyway we can just use the standard rule of adding two convergent series:
$$
sum_{n=1}^infty (a_n + b_n) = sum_{n=1}^infty a_n + sum_{n=1}^infty b_n.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. I assume by extension its also true for multiplying and division too? And also were $x$ is a complex number too ?
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:16






  • 2




    $begingroup$
    For division you'd need that the divisor doesn't converge to $0$. It also works for complex number. The proofs of these facts are pretty standard and shouldn't be too hard to find.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:17










  • $begingroup$
    Thanks again. I am just beginning in analysis as a self-interest. my background is CS.
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:18














2












2








2





$begingroup$

Yes, since your $x_0$ is fixed anyway we can just use the standard rule of adding two convergent series:
$$
sum_{n=1}^infty (a_n + b_n) = sum_{n=1}^infty a_n + sum_{n=1}^infty b_n.
$$






share|cite|improve this answer









$endgroup$



Yes, since your $x_0$ is fixed anyway we can just use the standard rule of adding two convergent series:
$$
sum_{n=1}^infty (a_n + b_n) = sum_{n=1}^infty a_n + sum_{n=1}^infty b_n.
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 14 '18 at 8:11









BigbearZzzBigbearZzz

8,88821652




8,88821652












  • $begingroup$
    Thank you. I assume by extension its also true for multiplying and division too? And also were $x$ is a complex number too ?
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:16






  • 2




    $begingroup$
    For division you'd need that the divisor doesn't converge to $0$. It also works for complex number. The proofs of these facts are pretty standard and shouldn't be too hard to find.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:17










  • $begingroup$
    Thanks again. I am just beginning in analysis as a self-interest. my background is CS.
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:18


















  • $begingroup$
    Thank you. I assume by extension its also true for multiplying and division too? And also were $x$ is a complex number too ?
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:16






  • 2




    $begingroup$
    For division you'd need that the divisor doesn't converge to $0$. It also works for complex number. The proofs of these facts are pretty standard and shouldn't be too hard to find.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:17










  • $begingroup$
    Thanks again. I am just beginning in analysis as a self-interest. my background is CS.
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:18
















$begingroup$
Thank you. I assume by extension its also true for multiplying and division too? And also were $x$ is a complex number too ?
$endgroup$
– TheoryQuest1
Dec 14 '18 at 8:16




$begingroup$
Thank you. I assume by extension its also true for multiplying and division too? And also were $x$ is a complex number too ?
$endgroup$
– TheoryQuest1
Dec 14 '18 at 8:16




2




2




$begingroup$
For division you'd need that the divisor doesn't converge to $0$. It also works for complex number. The proofs of these facts are pretty standard and shouldn't be too hard to find.
$endgroup$
– BigbearZzz
Dec 14 '18 at 8:17




$begingroup$
For division you'd need that the divisor doesn't converge to $0$. It also works for complex number. The proofs of these facts are pretty standard and shouldn't be too hard to find.
$endgroup$
– BigbearZzz
Dec 14 '18 at 8:17












$begingroup$
Thanks again. I am just beginning in analysis as a self-interest. my background is CS.
$endgroup$
– TheoryQuest1
Dec 14 '18 at 8:18




$begingroup$
Thanks again. I am just beginning in analysis as a self-interest. my background is CS.
$endgroup$
– TheoryQuest1
Dec 14 '18 at 8:18


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039092%2fcan-we-combine-two-convergent-infinite-series-as-follows%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