If $ sum a_n$ converges, does $sumlimits_{n=2}^{infty} frac{sqrt{a_n}}{ln n}cdot left( n^{a_n}-1 right)$...












4












$begingroup$



If $ sum a_n$ converges and $a_n>0$ for every $n$, does $sumlimits_{n=2}^{infty} frac{sqrt{a_n}}{ln n}cdot left( n^{a_n}-1 right)$ converge as well?




What I did: Define $u_n=frac{sqrt{a_n}}{ln n}cdot left( n^{a_n}-1 right)$. Trying to use the ratio test, I consider:
$$ frac{u_{n+1}}{u_n}=frac{ln(n)}{ln(n+1)} cdot sqrt{frac{a_{n+1}}{a_n}} cdotfrac{n^{a_{n+1}}-1}{n^{a_{n}}-1}=frac{ln(n)}{ln(n+1)} cdot sqrt{frac{a_{n+1}}{a_n}} cdotfrac{frac{n^{a_{n+1}}}{n^{a_{n}}}-frac{1}{n^{a_{n}}}}{1-frac{1}{n^{a_{n}}}}$$
and now:
$$ frac{ln(n)}{ln(n+1)} rightarrow 1 $$
$$ sqrt{frac{a_{n+1}}{a_n}} rightarrow g in (0;1) $$
but what can I do with the last ratio $$frac{frac{n^{a_{n+1}}}{n^{a_{n}}}-frac{1}{n^{a_{n}}}}{1-frac{1}{n^{a_{n}}}} ?$$
Or has somebody any other idea for this task?






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    We still can have $g=1$ and $sumlimits_{n=1} a_n < infty$ ...
    $endgroup$
    – rtybase
    Dec 17 '18 at 17:48
















4












$begingroup$



If $ sum a_n$ converges and $a_n>0$ for every $n$, does $sumlimits_{n=2}^{infty} frac{sqrt{a_n}}{ln n}cdot left( n^{a_n}-1 right)$ converge as well?




What I did: Define $u_n=frac{sqrt{a_n}}{ln n}cdot left( n^{a_n}-1 right)$. Trying to use the ratio test, I consider:
$$ frac{u_{n+1}}{u_n}=frac{ln(n)}{ln(n+1)} cdot sqrt{frac{a_{n+1}}{a_n}} cdotfrac{n^{a_{n+1}}-1}{n^{a_{n}}-1}=frac{ln(n)}{ln(n+1)} cdot sqrt{frac{a_{n+1}}{a_n}} cdotfrac{frac{n^{a_{n+1}}}{n^{a_{n}}}-frac{1}{n^{a_{n}}}}{1-frac{1}{n^{a_{n}}}}$$
and now:
$$ frac{ln(n)}{ln(n+1)} rightarrow 1 $$
$$ sqrt{frac{a_{n+1}}{a_n}} rightarrow g in (0;1) $$
but what can I do with the last ratio $$frac{frac{n^{a_{n+1}}}{n^{a_{n}}}-frac{1}{n^{a_{n}}}}{1-frac{1}{n^{a_{n}}}} ?$$
Or has somebody any other idea for this task?






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    We still can have $g=1$ and $sumlimits_{n=1} a_n < infty$ ...
    $endgroup$
    – rtybase
    Dec 17 '18 at 17:48














4












4








4


1



$begingroup$



If $ sum a_n$ converges and $a_n>0$ for every $n$, does $sumlimits_{n=2}^{infty} frac{sqrt{a_n}}{ln n}cdot left( n^{a_n}-1 right)$ converge as well?




What I did: Define $u_n=frac{sqrt{a_n}}{ln n}cdot left( n^{a_n}-1 right)$. Trying to use the ratio test, I consider:
$$ frac{u_{n+1}}{u_n}=frac{ln(n)}{ln(n+1)} cdot sqrt{frac{a_{n+1}}{a_n}} cdotfrac{n^{a_{n+1}}-1}{n^{a_{n}}-1}=frac{ln(n)}{ln(n+1)} cdot sqrt{frac{a_{n+1}}{a_n}} cdotfrac{frac{n^{a_{n+1}}}{n^{a_{n}}}-frac{1}{n^{a_{n}}}}{1-frac{1}{n^{a_{n}}}}$$
and now:
$$ frac{ln(n)}{ln(n+1)} rightarrow 1 $$
$$ sqrt{frac{a_{n+1}}{a_n}} rightarrow g in (0;1) $$
but what can I do with the last ratio $$frac{frac{n^{a_{n+1}}}{n^{a_{n}}}-frac{1}{n^{a_{n}}}}{1-frac{1}{n^{a_{n}}}} ?$$
Or has somebody any other idea for this task?






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$





If $ sum a_n$ converges and $a_n>0$ for every $n$, does $sumlimits_{n=2}^{infty} frac{sqrt{a_n}}{ln n}cdot left( n^{a_n}-1 right)$ converge as well?




What I did: Define $u_n=frac{sqrt{a_n}}{ln n}cdot left( n^{a_n}-1 right)$. Trying to use the ratio test, I consider:
$$ frac{u_{n+1}}{u_n}=frac{ln(n)}{ln(n+1)} cdot sqrt{frac{a_{n+1}}{a_n}} cdotfrac{n^{a_{n+1}}-1}{n^{a_{n}}-1}=frac{ln(n)}{ln(n+1)} cdot sqrt{frac{a_{n+1}}{a_n}} cdotfrac{frac{n^{a_{n+1}}}{n^{a_{n}}}-frac{1}{n^{a_{n}}}}{1-frac{1}{n^{a_{n}}}}$$
and now:
$$ frac{ln(n)}{ln(n+1)} rightarrow 1 $$
$$ sqrt{frac{a_{n+1}}{a_n}} rightarrow g in (0;1) $$
but what can I do with the last ratio $$frac{frac{n^{a_{n+1}}}{n^{a_{n}}}-frac{1}{n^{a_{n}}}}{1-frac{1}{n^{a_{n}}}} ?$$
Or has somebody any other idea for this task?






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 18:41









Did

248k23226465




248k23226465










asked Dec 17 '18 at 17:34









VirtualUserVirtualUser

1,096117




1,096117








  • 1




    $begingroup$
    We still can have $g=1$ and $sumlimits_{n=1} a_n < infty$ ...
    $endgroup$
    – rtybase
    Dec 17 '18 at 17:48














  • 1




    $begingroup$
    We still can have $g=1$ and $sumlimits_{n=1} a_n < infty$ ...
    $endgroup$
    – rtybase
    Dec 17 '18 at 17:48








1




1




$begingroup$
We still can have $g=1$ and $sumlimits_{n=1} a_n < infty$ ...
$endgroup$
– rtybase
Dec 17 '18 at 17:48




$begingroup$
We still can have $g=1$ and $sumlimits_{n=1} a_n < infty$ ...
$endgroup$
– rtybase
Dec 17 '18 at 17:48










1 Answer
1






active

oldest

votes


















5












$begingroup$

0. Note that the ratio test here is completely useless, because you do not know anything about $frac{a_{n+1}}{a_n}$ (it could even have a subsequence going to infinity).



1. Assume $a_n=O(log(n)^{-1})$. Then, $$frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)=(exp(log(n)a_n)-1)frac{a_n^{1/2}}{log(n)} leq Ca_n^{3/2} leq C’a_n$$ so the series converges.



2. Without the assumption that $a_n=O(log(n)^{-1})$, the series may diverge, see example below.



Take $a_{2^{p^4}}=p^{-2}$ (and zero everywhere else). Then, $sum_n{|a_n|}=sum_p{p^{-2}}$ converges.



However, $a_{2^{p^4}}log(2^{p^4}) =log(2) p^2 rightarrow infty$, thus $a_n neq O((log(n)^{-1})$.



Furthermore,
$$sum_n{frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)} = sum_p{frac{p^{-1}}{log(2)p^4}(2^{p^2}-1)}=infty$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I am afraid it is not. See the counter-example at the end.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 18:11










  • $begingroup$
    I am looking at this and I am not able to understand why we can assume that $ a_n=O(log(n)^{-1}) $ ?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 18:54










  • $begingroup$
    The point is that we cannot. If $a_n$ is $O(log(n)^{-1})$, then the second series converges. Else, it is possible that it does not converge.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:19












  • $begingroup$
    We cannot? But you done this in your proposition?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 20:21






  • 1




    $begingroup$
    I thought I had made myself clear. I proved two things: 1) in general the series needs not converge. 2) if $a_n$ is $O(log(n)^{-1})$ then the series converges.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:34











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%2f3044218%2fif-sum-a-n-converges-does-sum-limits-n-2-infty-frac-sqrta-n-ln%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









5












$begingroup$

0. Note that the ratio test here is completely useless, because you do not know anything about $frac{a_{n+1}}{a_n}$ (it could even have a subsequence going to infinity).



1. Assume $a_n=O(log(n)^{-1})$. Then, $$frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)=(exp(log(n)a_n)-1)frac{a_n^{1/2}}{log(n)} leq Ca_n^{3/2} leq C’a_n$$ so the series converges.



2. Without the assumption that $a_n=O(log(n)^{-1})$, the series may diverge, see example below.



Take $a_{2^{p^4}}=p^{-2}$ (and zero everywhere else). Then, $sum_n{|a_n|}=sum_p{p^{-2}}$ converges.



However, $a_{2^{p^4}}log(2^{p^4}) =log(2) p^2 rightarrow infty$, thus $a_n neq O((log(n)^{-1})$.



Furthermore,
$$sum_n{frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)} = sum_p{frac{p^{-1}}{log(2)p^4}(2^{p^2}-1)}=infty$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I am afraid it is not. See the counter-example at the end.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 18:11










  • $begingroup$
    I am looking at this and I am not able to understand why we can assume that $ a_n=O(log(n)^{-1}) $ ?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 18:54










  • $begingroup$
    The point is that we cannot. If $a_n$ is $O(log(n)^{-1})$, then the second series converges. Else, it is possible that it does not converge.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:19












  • $begingroup$
    We cannot? But you done this in your proposition?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 20:21






  • 1




    $begingroup$
    I thought I had made myself clear. I proved two things: 1) in general the series needs not converge. 2) if $a_n$ is $O(log(n)^{-1})$ then the series converges.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:34
















5












$begingroup$

0. Note that the ratio test here is completely useless, because you do not know anything about $frac{a_{n+1}}{a_n}$ (it could even have a subsequence going to infinity).



1. Assume $a_n=O(log(n)^{-1})$. Then, $$frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)=(exp(log(n)a_n)-1)frac{a_n^{1/2}}{log(n)} leq Ca_n^{3/2} leq C’a_n$$ so the series converges.



2. Without the assumption that $a_n=O(log(n)^{-1})$, the series may diverge, see example below.



Take $a_{2^{p^4}}=p^{-2}$ (and zero everywhere else). Then, $sum_n{|a_n|}=sum_p{p^{-2}}$ converges.



However, $a_{2^{p^4}}log(2^{p^4}) =log(2) p^2 rightarrow infty$, thus $a_n neq O((log(n)^{-1})$.



Furthermore,
$$sum_n{frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)} = sum_p{frac{p^{-1}}{log(2)p^4}(2^{p^2}-1)}=infty$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I am afraid it is not. See the counter-example at the end.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 18:11










  • $begingroup$
    I am looking at this and I am not able to understand why we can assume that $ a_n=O(log(n)^{-1}) $ ?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 18:54










  • $begingroup$
    The point is that we cannot. If $a_n$ is $O(log(n)^{-1})$, then the second series converges. Else, it is possible that it does not converge.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:19












  • $begingroup$
    We cannot? But you done this in your proposition?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 20:21






  • 1




    $begingroup$
    I thought I had made myself clear. I proved two things: 1) in general the series needs not converge. 2) if $a_n$ is $O(log(n)^{-1})$ then the series converges.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:34














5












5








5





$begingroup$

0. Note that the ratio test here is completely useless, because you do not know anything about $frac{a_{n+1}}{a_n}$ (it could even have a subsequence going to infinity).



1. Assume $a_n=O(log(n)^{-1})$. Then, $$frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)=(exp(log(n)a_n)-1)frac{a_n^{1/2}}{log(n)} leq Ca_n^{3/2} leq C’a_n$$ so the series converges.



2. Without the assumption that $a_n=O(log(n)^{-1})$, the series may diverge, see example below.



Take $a_{2^{p^4}}=p^{-2}$ (and zero everywhere else). Then, $sum_n{|a_n|}=sum_p{p^{-2}}$ converges.



However, $a_{2^{p^4}}log(2^{p^4}) =log(2) p^2 rightarrow infty$, thus $a_n neq O((log(n)^{-1})$.



Furthermore,
$$sum_n{frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)} = sum_p{frac{p^{-1}}{log(2)p^4}(2^{p^2}-1)}=infty$$






share|cite|improve this answer











$endgroup$



0. Note that the ratio test here is completely useless, because you do not know anything about $frac{a_{n+1}}{a_n}$ (it could even have a subsequence going to infinity).



1. Assume $a_n=O(log(n)^{-1})$. Then, $$frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)=(exp(log(n)a_n)-1)frac{a_n^{1/2}}{log(n)} leq Ca_n^{3/2} leq C’a_n$$ so the series converges.



2. Without the assumption that $a_n=O(log(n)^{-1})$, the series may diverge, see example below.



Take $a_{2^{p^4}}=p^{-2}$ (and zero everywhere else). Then, $sum_n{|a_n|}=sum_p{p^{-2}}$ converges.



However, $a_{2^{p^4}}log(2^{p^4}) =log(2) p^2 rightarrow infty$, thus $a_n neq O((log(n)^{-1})$.



Furthermore,
$$sum_n{frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)} = sum_p{frac{p^{-1}}{log(2)p^4}(2^{p^2}-1)}=infty$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 18 '18 at 11:12









Did

248k23226465




248k23226465










answered Dec 17 '18 at 18:08









MindlackMindlack

4,920211




4,920211








  • 1




    $begingroup$
    I am afraid it is not. See the counter-example at the end.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 18:11










  • $begingroup$
    I am looking at this and I am not able to understand why we can assume that $ a_n=O(log(n)^{-1}) $ ?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 18:54










  • $begingroup$
    The point is that we cannot. If $a_n$ is $O(log(n)^{-1})$, then the second series converges. Else, it is possible that it does not converge.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:19












  • $begingroup$
    We cannot? But you done this in your proposition?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 20:21






  • 1




    $begingroup$
    I thought I had made myself clear. I proved two things: 1) in general the series needs not converge. 2) if $a_n$ is $O(log(n)^{-1})$ then the series converges.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:34














  • 1




    $begingroup$
    I am afraid it is not. See the counter-example at the end.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 18:11










  • $begingroup$
    I am looking at this and I am not able to understand why we can assume that $ a_n=O(log(n)^{-1}) $ ?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 18:54










  • $begingroup$
    The point is that we cannot. If $a_n$ is $O(log(n)^{-1})$, then the second series converges. Else, it is possible that it does not converge.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:19












  • $begingroup$
    We cannot? But you done this in your proposition?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 20:21






  • 1




    $begingroup$
    I thought I had made myself clear. I proved two things: 1) in general the series needs not converge. 2) if $a_n$ is $O(log(n)^{-1})$ then the series converges.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:34








1




1




$begingroup$
I am afraid it is not. See the counter-example at the end.
$endgroup$
– Mindlack
Dec 17 '18 at 18:11




$begingroup$
I am afraid it is not. See the counter-example at the end.
$endgroup$
– Mindlack
Dec 17 '18 at 18:11












$begingroup$
I am looking at this and I am not able to understand why we can assume that $ a_n=O(log(n)^{-1}) $ ?
$endgroup$
– VirtualUser
Dec 17 '18 at 18:54




$begingroup$
I am looking at this and I am not able to understand why we can assume that $ a_n=O(log(n)^{-1}) $ ?
$endgroup$
– VirtualUser
Dec 17 '18 at 18:54












$begingroup$
The point is that we cannot. If $a_n$ is $O(log(n)^{-1})$, then the second series converges. Else, it is possible that it does not converge.
$endgroup$
– Mindlack
Dec 17 '18 at 20:19






$begingroup$
The point is that we cannot. If $a_n$ is $O(log(n)^{-1})$, then the second series converges. Else, it is possible that it does not converge.
$endgroup$
– Mindlack
Dec 17 '18 at 20:19














$begingroup$
We cannot? But you done this in your proposition?
$endgroup$
– VirtualUser
Dec 17 '18 at 20:21




$begingroup$
We cannot? But you done this in your proposition?
$endgroup$
– VirtualUser
Dec 17 '18 at 20:21




1




1




$begingroup$
I thought I had made myself clear. I proved two things: 1) in general the series needs not converge. 2) if $a_n$ is $O(log(n)^{-1})$ then the series converges.
$endgroup$
– Mindlack
Dec 17 '18 at 20:34




$begingroup$
I thought I had made myself clear. I proved two things: 1) in general the series needs not converge. 2) if $a_n$ is $O(log(n)^{-1})$ then the series converges.
$endgroup$
– Mindlack
Dec 17 '18 at 20:34


















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%2f3044218%2fif-sum-a-n-converges-does-sum-limits-n-2-infty-frac-sqrta-n-ln%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