sum of $sum_{n=1}^{infty}(-1)^{n+1}(2^{1/n}-1)$
up vote
-1
down vote
favorite
Is there any way to analytically find the sum of an infinite series if the infinite series is not geometric, nor can it be set up as residues for a convenient meromorphic function?
For instance, if I have the function
$f(x)=1+sum_{n=1}^{infty}(-1)^{n+1}(x^{1/n}-1)$, I have a function of x that will converge conditionally for all $xgeq1$, because it is an alternating series where the limit of terms converges to zero. However, I don't see any way of solving the sum using residues.
Are there any other methods to attempt to find the value of the series (and therefore of $f(x)$)?
sequences-and-series
edited Nov 21 at 11:22
Tianlalu
3,01021038
asked Nov 21 at 11:14
MoKo19
1914
add a comment |
up vote
-1
down vote
favorite
Is there any way to analytically find the sum of an infinite series if the infinite series is not geometric, nor can it be set up as residues for a convenient meromorphic function?
For instance, if I have the function
$f(x)=1+sum_{n=1}^{infty}(-1)^{n+1}(x^{1/n}-1)$, I have a function of x that will converge conditionally for all $xgeq1$, because it is an alternating series where the limit of terms converges to zero. However, I don't see any way of solving the sum using residues.
Are there any other methods to attempt to find the value of the series (and therefore of $f(x)$)?
sequences-and-series
edited Nov 21 at 11:22
Tianlalu
3,01021038
asked Nov 21 at 11:14
MoKo19
1914
You can definitely approximate it using the fact that $$lim_{x to 0} left( frac{a^x - 1}{x} right) = ln{a}, qquad forall~a > 0$$
– rtybase
Nov 21 at 13:36
@rtybase: How does that help?
– MoKo19
Nov 21 at 13:46
E.g. $$sumlimits_{n=1}^{infty}(-1)^{n+1}(2^{1/n}-1) approx sum_{n=1}^{N(varepsilon)}(-1)^{n+1}(2^{1/n}-1)+ln{2}left(sum_{n=N(varepsilon)+1}^{infty}(-1)^{n+1}frac{1}{n}right)$$
– rtybase
Nov 21 at 14:02
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Is there any way to analytically find the sum of an infinite series if the infinite series is not geometric, nor can it be set up as residues for a convenient meromorphic function?
For instance, if I have the function
$f(x)=1+sum_{n=1}^{infty}(-1)^{n+1}(x^{1/n}-1)$, I have a function of x that will converge conditionally for all $xgeq1$, because it is an alternating series where the limit of terms converges to zero. However, I don't see any way of solving the sum using residues.
Are there any other methods to attempt to find the value of the series (and therefore of $f(x)$)?
sequences-and-series
edited Nov 21 at 11:22
Tianlalu
3,01021038
asked Nov 21 at 11:14
MoKo19
1914
Is there any way to analytically find the sum of an infinite series if the infinite series is not geometric, nor can it be set up as residues for a convenient meromorphic function?
For instance, if I have the function
$f(x)=1+sum_{n=1}^{infty}(-1)^{n+1}(x^{1/n}-1)$, I have a function of x that will converge conditionally for all $xgeq1$, because it is an alternating series where the limit of terms converges to zero. However, I don't see any way of solving the sum using residues.
Are there any other methods to attempt to find the value of the series (and therefore of $f(x)$)?
sequences-and-series
sequences-and-series
edited Nov 21 at 11:22
Tianlalu
3,01021038
asked Nov 21 at 11:14
MoKo19
1914
edited Nov 21 at 11:22
Tianlalu
3,01021038
asked Nov 21 at 11:14
MoKo19
1914
edited Nov 21 at 11:22
Tianlalu
3,01021038
edited Nov 21 at 11:22
Tianlalu
3,01021038
edited Nov 21 at 11:22
Tianlalu
3,01021038
3,01021038
asked Nov 21 at 11:14
MoKo19
1914
asked Nov 21 at 11:14
MoKo19
1914
asked Nov 21 at 11:14
MoKo19
1914
1914
You can definitely approximate it using the fact that $$lim_{x to 0} left( frac{a^x - 1}{x} right) = ln{a}, qquad forall~a > 0$$
– rtybase
Nov 21 at 13:36
@rtybase: How does that help?
– MoKo19
Nov 21 at 13:46
E.g. $$sumlimits_{n=1}^{infty}(-1)^{n+1}(2^{1/n}-1) approx sum_{n=1}^{N(varepsilon)}(-1)^{n+1}(2^{1/n}-1)+ln{2}left(sum_{n=N(varepsilon)+1}^{infty}(-1)^{n+1}frac{1}{n}right)$$
– rtybase
Nov 21 at 14:02
add a comment |
You can definitely approximate it using the fact that $$lim_{x to 0} left( frac{a^x - 1}{x} right) = ln{a}, qquad forall~a > 0$$
– rtybase
Nov 21 at 13:36
@rtybase: How does that help?
– MoKo19
Nov 21 at 13:46
E.g. $$sumlimits_{n=1}^{infty}(-1)^{n+1}(2^{1/n}-1) approx sum_{n=1}^{N(varepsilon)}(-1)^{n+1}(2^{1/n}-1)+ln{2}left(sum_{n=N(varepsilon)+1}^{infty}(-1)^{n+1}frac{1}{n}right)$$
– rtybase
Nov 21 at 14:02
You can definitely approximate it using the fact that $$lim_{x to 0} left( frac{a^x - 1}{x} right) = ln{a}, qquad forall~a > 0$$
– rtybase
Nov 21 at 13:36
You can definitely approximate it using the fact that $$lim_{x to 0} left( frac{a^x - 1}{x} right) = ln{a}, qquad forall~a > 0$$
– rtybase
Nov 21 at 13:36
@rtybase: How does that help?
– MoKo19
Nov 21 at 13:46
@rtybase: How does that help?
– MoKo19
Nov 21 at 13:46
E.g. $$sumlimits_{n=1}^{infty}(-1)^{n+1}(2^{1/n}-1) approx sum_{n=1}^{N(varepsilon)}(-1)^{n+1}(2^{1/n}-1)+ln{2}left(sum_{n=N(varepsilon)+1}^{infty}(-1)^{n+1}frac{1}{n}right)$$
– rtybase
Nov 21 at 14:02
E.g. $$sumlimits_{n=1}^{infty}(-1)^{n+1}(2^{1/n}-1) approx sum_{n=1}^{N(varepsilon)}(-1)^{n+1}(2^{1/n}-1)+ln{2}left(sum_{n=N(varepsilon)+1}^{infty}(-1)^{n+1}frac{1}{n}right)$$
– rtybase
Nov 21 at 14:02
add a comment |
active
oldest
votes
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007575%2fsum-of-sum-n-1-infty-1n121-n-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007575%2fsum-of-sum-n-1-infty-1n121-n-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
You can definitely approximate it using the fact that $$lim_{x to 0} left( frac{a^x - 1}{x} right) = ln{a}, qquad forall~a > 0$$
– rtybase
Nov 21 at 13:36
@rtybase: How does that help?
– MoKo19
Nov 21 at 13:46
E.g. $$sumlimits_{n=1}^{infty}(-1)^{n+1}(2^{1/n}-1) approx sum_{n=1}^{N(varepsilon)}(-1)^{n+1}(2^{1/n}-1)+ln{2}left(sum_{n=N(varepsilon)+1}^{infty}(-1)^{n+1}frac{1}{n}right)$$
– rtybase
Nov 21 at 14:02