Evaluation of this infinite series












0












$begingroup$


Given the following infinite series:



$sum_{n=1}^{infty}frac{1}{(n)^{p}}-frac{1}{(n+1)^{p}}$



Find the summation of this series?



My approach:



The above series expands to:



Sum = $(frac{1}{1^p}-frac{1}{2^p}) +(frac{1}{2^p}-frac{1}{3^p}) + (frac{1}{3^p}-frac{1}{4^p})+ (frac{1}{4^p}-frac{1}{5^p}).... (frac{1}{(infty)^p}-frac{1}{(infty+2)^p})$



Since, from second term onward, all neighboring terms cancel each other out all the 'middle' terms will cancel each other out. Thus it will evaluate to:



Sum = $(frac{1}{1^p} - frac{1}{(infty+2)^p})$



As, for any finite real/complex value of: $p$ $(infty+2)^p = infty$. Thus,



Sum = $(frac{1}{1^p} - frac{1}{infty})$.



As, $frac{1}{infty} = 0$. Thus:



Sum = $frac{1}{1^p}$. (for any finite value of $p$).



Query: Is the above argument correct and can we can replace the series with $frac{1}{1^p}$ in any equation or not?






real-analysis







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    You're throwing around $infty$ like it's an actual number...
    $endgroup$
    – Andrew Li
    Dec 19 '18 at 12:35










  • $begingroup$
    No it is not sound. Fortunately there is a very simple way to make it rigorous.
    $endgroup$
    – Did
    Dec 19 '18 at 12:38










  • $begingroup$
    @Did I was wondering if the method below (in the answer) is what you had in mind when you said that there is a simple way to make it rigorous?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:59






  • 2




    $begingroup$
    It is. $ $ $ $ $ $
    $endgroup$
    – Did
    Dec 19 '18 at 13:04
















0












$begingroup$


Given the following infinite series:



$sum_{n=1}^{infty}frac{1}{(n)^{p}}-frac{1}{(n+1)^{p}}$



Find the summation of this series?



My approach:



The above series expands to:



Sum = $(frac{1}{1^p}-frac{1}{2^p}) +(frac{1}{2^p}-frac{1}{3^p}) + (frac{1}{3^p}-frac{1}{4^p})+ (frac{1}{4^p}-frac{1}{5^p}).... (frac{1}{(infty)^p}-frac{1}{(infty+2)^p})$



Since, from second term onward, all neighboring terms cancel each other out all the 'middle' terms will cancel each other out. Thus it will evaluate to:



Sum = $(frac{1}{1^p} - frac{1}{(infty+2)^p})$



As, for any finite real/complex value of: $p$ $(infty+2)^p = infty$. Thus,



Sum = $(frac{1}{1^p} - frac{1}{infty})$.



As, $frac{1}{infty} = 0$. Thus:



Sum = $frac{1}{1^p}$. (for any finite value of $p$).



Query: Is the above argument correct and can we can replace the series with $frac{1}{1^p}$ in any equation or not?






real-analysis







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    You're throwing around $infty$ like it's an actual number...
    $endgroup$
    – Andrew Li
    Dec 19 '18 at 12:35










  • $begingroup$
    No it is not sound. Fortunately there is a very simple way to make it rigorous.
    $endgroup$
    – Did
    Dec 19 '18 at 12:38










  • $begingroup$
    @Did I was wondering if the method below (in the answer) is what you had in mind when you said that there is a simple way to make it rigorous?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:59






  • 2




    $begingroup$
    It is. $ $ $ $ $ $
    $endgroup$
    – Did
    Dec 19 '18 at 13:04














0












0








0





$begingroup$


Given the following infinite series:



$sum_{n=1}^{infty}frac{1}{(n)^{p}}-frac{1}{(n+1)^{p}}$



Find the summation of this series?



My approach:



The above series expands to:



Sum = $(frac{1}{1^p}-frac{1}{2^p}) +(frac{1}{2^p}-frac{1}{3^p}) + (frac{1}{3^p}-frac{1}{4^p})+ (frac{1}{4^p}-frac{1}{5^p}).... (frac{1}{(infty)^p}-frac{1}{(infty+2)^p})$



Since, from second term onward, all neighboring terms cancel each other out all the 'middle' terms will cancel each other out. Thus it will evaluate to:



Sum = $(frac{1}{1^p} - frac{1}{(infty+2)^p})$



As, for any finite real/complex value of: $p$ $(infty+2)^p = infty$. Thus,



Sum = $(frac{1}{1^p} - frac{1}{infty})$.



As, $frac{1}{infty} = 0$. Thus:



Sum = $frac{1}{1^p}$. (for any finite value of $p$).



Query: Is the above argument correct and can we can replace the series with $frac{1}{1^p}$ in any equation or not?






real-analysis







share|cite|improve this question









$endgroup$




Given the following infinite series:



$sum_{n=1}^{infty}frac{1}{(n)^{p}}-frac{1}{(n+1)^{p}}$



Find the summation of this series?



My approach:



The above series expands to:



Sum = $(frac{1}{1^p}-frac{1}{2^p}) +(frac{1}{2^p}-frac{1}{3^p}) + (frac{1}{3^p}-frac{1}{4^p})+ (frac{1}{4^p}-frac{1}{5^p}).... (frac{1}{(infty)^p}-frac{1}{(infty+2)^p})$



Since, from second term onward, all neighboring terms cancel each other out all the 'middle' terms will cancel each other out. Thus it will evaluate to:



Sum = $(frac{1}{1^p} - frac{1}{(infty+2)^p})$



As, for any finite real/complex value of: $p$ $(infty+2)^p = infty$. Thus,



Sum = $(frac{1}{1^p} - frac{1}{infty})$.



As, $frac{1}{infty} = 0$. Thus:



Sum = $frac{1}{1^p}$. (for any finite value of $p$).



Query: Is the above argument correct and can we can replace the series with $frac{1}{1^p}$ in any equation or not?






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 19 '18 at 12:33









TheoryQuest1TheoryQuest1

1498




1498








  • 2




    $begingroup$
    You're throwing around $infty$ like it's an actual number...
    $endgroup$
    – Andrew Li
    Dec 19 '18 at 12:35










  • $begingroup$
    No it is not sound. Fortunately there is a very simple way to make it rigorous.
    $endgroup$
    – Did
    Dec 19 '18 at 12:38










  • $begingroup$
    @Did I was wondering if the method below (in the answer) is what you had in mind when you said that there is a simple way to make it rigorous?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:59






  • 2




    $begingroup$
    It is. $ $ $ $ $ $
    $endgroup$
    – Did
    Dec 19 '18 at 13:04














  • 2




    $begingroup$
    You're throwing around $infty$ like it's an actual number...
    $endgroup$
    – Andrew Li
    Dec 19 '18 at 12:35










  • $begingroup$
    No it is not sound. Fortunately there is a very simple way to make it rigorous.
    $endgroup$
    – Did
    Dec 19 '18 at 12:38










  • $begingroup$
    @Did I was wondering if the method below (in the answer) is what you had in mind when you said that there is a simple way to make it rigorous?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:59






  • 2




    $begingroup$
    It is. $ $ $ $ $ $
    $endgroup$
    – Did
    Dec 19 '18 at 13:04








2




2




$begingroup$
You're throwing around $infty$ like it's an actual number...
$endgroup$
– Andrew Li
Dec 19 '18 at 12:35




$begingroup$
You're throwing around $infty$ like it's an actual number...
$endgroup$
– Andrew Li
Dec 19 '18 at 12:35












$begingroup$
No it is not sound. Fortunately there is a very simple way to make it rigorous.
$endgroup$
– Did
Dec 19 '18 at 12:38




$begingroup$
No it is not sound. Fortunately there is a very simple way to make it rigorous.
$endgroup$
– Did
Dec 19 '18 at 12:38












$begingroup$
@Did I was wondering if the method below (in the answer) is what you had in mind when you said that there is a simple way to make it rigorous?
$endgroup$
– TheoryQuest1
Dec 19 '18 at 12:59




$begingroup$
@Did I was wondering if the method below (in the answer) is what you had in mind when you said that there is a simple way to make it rigorous?
$endgroup$
– TheoryQuest1
Dec 19 '18 at 12:59




2




2




$begingroup$
It is. $ $ $ $ $ $
$endgroup$
– Did
Dec 19 '18 at 13:04




$begingroup$
It is. $ $ $ $ $ $
$endgroup$
– Did
Dec 19 '18 at 13:04










1 Answer
1






active

oldest

votes


















4












$begingroup$

Yes you can, although the symbol $infty$ can be handled in a different way. For example, call



$$
S_N = sum_{n=1}^Nfrac{1}{n^p} - frac{1}{(n + 1)^p}
$$



You showed that



$$
S_N = 1 - frac{1}{(N + 1)^p}
$$



The sum you are interested in is



$$
S = lim_{N to infty} S_N = 1
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. As I understand, when we say $S = lim_{N to infty} S_N = 1$ instead of $frac{1}{1^p}$ we essentially imply that as $lim_{N to infty} then frac{1}{(N + 1)^p} to 0$. Thus, it is never exactly 0 but goes asymptotically to 0. Now, here is what I am struggling with. (1) If I remember correctly, $.99999...$ with infinite digits is considered equal to 1. Why can't we have the same argument for this case?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:52












  • $begingroup$
    (2) In any equation, do we replace the series, with $S = lim_{N to infty} S_N = 1$ as the sum instead of 1 (as I thought) ?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:53








  • 1




    $begingroup$
    @TheoryQuest1 You can replace the series as 1, as it converges to 1. You can see this other post, maybe it will help you
    $endgroup$
    – caverac
    Dec 19 '18 at 12:56












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%2f3046334%2fevaluation-of-this-infinite-series%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









4












$begingroup$

Yes you can, although the symbol $infty$ can be handled in a different way. For example, call



$$
S_N = sum_{n=1}^Nfrac{1}{n^p} - frac{1}{(n + 1)^p}
$$



You showed that



$$
S_N = 1 - frac{1}{(N + 1)^p}
$$



The sum you are interested in is



$$
S = lim_{N to infty} S_N = 1
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. As I understand, when we say $S = lim_{N to infty} S_N = 1$ instead of $frac{1}{1^p}$ we essentially imply that as $lim_{N to infty} then frac{1}{(N + 1)^p} to 0$. Thus, it is never exactly 0 but goes asymptotically to 0. Now, here is what I am struggling with. (1) If I remember correctly, $.99999...$ with infinite digits is considered equal to 1. Why can't we have the same argument for this case?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:52












  • $begingroup$
    (2) In any equation, do we replace the series, with $S = lim_{N to infty} S_N = 1$ as the sum instead of 1 (as I thought) ?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:53








  • 1




    $begingroup$
    @TheoryQuest1 You can replace the series as 1, as it converges to 1. You can see this other post, maybe it will help you
    $endgroup$
    – caverac
    Dec 19 '18 at 12:56
















4












$begingroup$

Yes you can, although the symbol $infty$ can be handled in a different way. For example, call



$$
S_N = sum_{n=1}^Nfrac{1}{n^p} - frac{1}{(n + 1)^p}
$$



You showed that



$$
S_N = 1 - frac{1}{(N + 1)^p}
$$



The sum you are interested in is



$$
S = lim_{N to infty} S_N = 1
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. As I understand, when we say $S = lim_{N to infty} S_N = 1$ instead of $frac{1}{1^p}$ we essentially imply that as $lim_{N to infty} then frac{1}{(N + 1)^p} to 0$. Thus, it is never exactly 0 but goes asymptotically to 0. Now, here is what I am struggling with. (1) If I remember correctly, $.99999...$ with infinite digits is considered equal to 1. Why can't we have the same argument for this case?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:52












  • $begingroup$
    (2) In any equation, do we replace the series, with $S = lim_{N to infty} S_N = 1$ as the sum instead of 1 (as I thought) ?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:53








  • 1




    $begingroup$
    @TheoryQuest1 You can replace the series as 1, as it converges to 1. You can see this other post, maybe it will help you
    $endgroup$
    – caverac
    Dec 19 '18 at 12:56














4












4








4





$begingroup$

Yes you can, although the symbol $infty$ can be handled in a different way. For example, call



$$
S_N = sum_{n=1}^Nfrac{1}{n^p} - frac{1}{(n + 1)^p}
$$



You showed that



$$
S_N = 1 - frac{1}{(N + 1)^p}
$$



The sum you are interested in is



$$
S = lim_{N to infty} S_N = 1
$$






share|cite|improve this answer









$endgroup$



Yes you can, although the symbol $infty$ can be handled in a different way. For example, call



$$
S_N = sum_{n=1}^Nfrac{1}{n^p} - frac{1}{(n + 1)^p}
$$



You showed that



$$
S_N = 1 - frac{1}{(N + 1)^p}
$$



The sum you are interested in is



$$
S = lim_{N to infty} S_N = 1
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 12:37









caveraccaverac

14.8k31130




14.8k31130












  • $begingroup$
    Thank you. As I understand, when we say $S = lim_{N to infty} S_N = 1$ instead of $frac{1}{1^p}$ we essentially imply that as $lim_{N to infty} then frac{1}{(N + 1)^p} to 0$. Thus, it is never exactly 0 but goes asymptotically to 0. Now, here is what I am struggling with. (1) If I remember correctly, $.99999...$ with infinite digits is considered equal to 1. Why can't we have the same argument for this case?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:52












  • $begingroup$
    (2) In any equation, do we replace the series, with $S = lim_{N to infty} S_N = 1$ as the sum instead of 1 (as I thought) ?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:53








  • 1




    $begingroup$
    @TheoryQuest1 You can replace the series as 1, as it converges to 1. You can see this other post, maybe it will help you
    $endgroup$
    – caverac
    Dec 19 '18 at 12:56


















  • $begingroup$
    Thank you. As I understand, when we say $S = lim_{N to infty} S_N = 1$ instead of $frac{1}{1^p}$ we essentially imply that as $lim_{N to infty} then frac{1}{(N + 1)^p} to 0$. Thus, it is never exactly 0 but goes asymptotically to 0. Now, here is what I am struggling with. (1) If I remember correctly, $.99999...$ with infinite digits is considered equal to 1. Why can't we have the same argument for this case?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:52












  • $begingroup$
    (2) In any equation, do we replace the series, with $S = lim_{N to infty} S_N = 1$ as the sum instead of 1 (as I thought) ?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:53








  • 1




    $begingroup$
    @TheoryQuest1 You can replace the series as 1, as it converges to 1. You can see this other post, maybe it will help you
    $endgroup$
    – caverac
    Dec 19 '18 at 12:56
















$begingroup$
Thank you. As I understand, when we say $S = lim_{N to infty} S_N = 1$ instead of $frac{1}{1^p}$ we essentially imply that as $lim_{N to infty} then frac{1}{(N + 1)^p} to 0$. Thus, it is never exactly 0 but goes asymptotically to 0. Now, here is what I am struggling with. (1) If I remember correctly, $.99999...$ with infinite digits is considered equal to 1. Why can't we have the same argument for this case?
$endgroup$
– TheoryQuest1
Dec 19 '18 at 12:52






$begingroup$
Thank you. As I understand, when we say $S = lim_{N to infty} S_N = 1$ instead of $frac{1}{1^p}$ we essentially imply that as $lim_{N to infty} then frac{1}{(N + 1)^p} to 0$. Thus, it is never exactly 0 but goes asymptotically to 0. Now, here is what I am struggling with. (1) If I remember correctly, $.99999...$ with infinite digits is considered equal to 1. Why can't we have the same argument for this case?
$endgroup$
– TheoryQuest1
Dec 19 '18 at 12:52














$begingroup$
(2) In any equation, do we replace the series, with $S = lim_{N to infty} S_N = 1$ as the sum instead of 1 (as I thought) ?
$endgroup$
– TheoryQuest1
Dec 19 '18 at 12:53






$begingroup$
(2) In any equation, do we replace the series, with $S = lim_{N to infty} S_N = 1$ as the sum instead of 1 (as I thought) ?
$endgroup$
– TheoryQuest1
Dec 19 '18 at 12:53






1




1




$begingroup$
@TheoryQuest1 You can replace the series as 1, as it converges to 1. You can see this other post, maybe it will help you
$endgroup$
– caverac
Dec 19 '18 at 12:56




$begingroup$
@TheoryQuest1 You can replace the series as 1, as it converges to 1. You can see this other post, maybe it will help you
$endgroup$
– caverac
Dec 19 '18 at 12:56


















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%2f3046334%2fevaluation-of-this-infinite-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

In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
Read more

How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

Image
1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
Read more