How do you determine if the following series converges?Convergence and exponentialsDetermine if the series...
How do I narratively explain how in-game circumstances do not mechanically allow a PC to instantly kill an NPC?
What is an explicit bijection in combinatorics?
Minimum Viable Product for RTS game?
Do the speed limit reductions due to pollution also apply to electric cars in France?
Coworker asking me to not bring cakes due to self control issue. What should I do?
Piping Multiple Numbers into Sed
Disk space full during insert, what happens?
Coworker is trying to get me to sign his petition to run for office. How to decline politely?
bash aliases do not expand even with shopt expand_aliases
Can someone explain European graduate programs in STEM fields?
Boss asked me to sign a resignation paper without a date on it along with my new contract
Does しかたない imply disappointment?
What is the reward?
Why do single electrical receptacles exist?
If I tried and failed to start my own business, how do I apply for a job without job experience?
Was the Spartan by Mimic Systems a real product?
How do I make my single-minded character more interested in the main story?
Short story about a man betting a group he could tell a story, and one of them would disappear and the others would not notice
How can I handle players killing my NPC outside of combat?
How do I fight with Heavy Armor as a Wizard with Tenser's Transformation?
Is there redundancy between a US Passport Card and an Enhanced Driver's License?
How can I deal with my coworker having zero social cues?
How do I add a strong "onion flavor" to the biryani (in restaurant style)?
Is Screenshot Time-tracking Common?
How do you determine if the following series converges?
Convergence and exponentialsDetermine if the series converges/divergesHow to prove this series converges $sum_{n=2}^infty frac{ln(n)}{n^{3/2}}$?How to determine if this series converges?Divergence test for the series $sum_{n=1}^{infty} (sqrt[n]{2n^2}-1)^n$Determine whether the series converges or diverges?My approach to determine if the following series is convergentHow to prove that $sum_{n=1}^{infty} frac{(log (n))^2}{n^2}$ converges?Determine whether the series converges or diverges.Does $displaystylesum_{n=1}^{infty}sinleft(displaystylefrac{1}{sqrt{n}}right)$ converge?Find all values of $k$ such that the series with terms $k^n / n^k$ converges.
$begingroup$
$$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{2}}$$
I tried using the limit comparison test with $$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{}}$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrm{e}^x$, but I don't know where else to start. Any suggestions?
convergence
asked 45 mins ago
JayJay
334
$endgroup$
add a comment |
$begingroup$
$$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{2}}$$
I tried using the limit comparison test with $$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{}}$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrm{e}^x$, but I don't know where else to start. Any suggestions?
convergence
asked 45 mins ago
JayJay
334
$endgroup$
$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
35 mins ago
$begingroup$
Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
$endgroup$
– JavaMan
29 mins ago
add a comment |
$begingroup$
$$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{2}}$$
I tried using the limit comparison test with $$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{}}$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrm{e}^x$, but I don't know where else to start. Any suggestions?
convergence
asked 45 mins ago
JayJay
334
$endgroup$
$$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{2}}$$
I tried using the limit comparison test with $$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{}}$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrm{e}^x$, but I don't know where else to start. Any suggestions?
convergence
convergence
asked 45 mins ago
JayJay
334
asked 45 mins ago
JayJay
334
asked 45 mins ago
JayJay
334
asked 45 mins ago
JayJay
334
asked 45 mins ago
JayJay
334
334
$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
35 mins ago
$begingroup$
Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
$endgroup$
– JavaMan
29 mins ago
add a comment |
$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
35 mins ago
$begingroup$
Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
$endgroup$
– JavaMan
29 mins ago
$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
35 mins ago
$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
35 mins ago
$begingroup$
Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
$endgroup$
– JavaMan
29 mins ago
$begingroup$
Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
$endgroup$
– JavaMan
29 mins ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
HINT:
Note that $$left( 1-frac1k right)^kle e^{-1}$$
Can you finish?
answered 30 mins ago
Mark ViolaMark Viola
132k1277174
$endgroup$
add a comment |
$begingroup$
The root test works. Consider
$$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
hence the series converges.
answered 34 mins ago
Theo BenditTheo Bendit
18.7k12253
$endgroup$
add a comment |
$begingroup$
The ratio test is also interesting
$$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
$$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$
Develop as a Taylor series for large values of $k$ to get
$$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
Continue with Taylor
$$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$
answered 21 mins ago
Claude LeiboviciClaude Leibovici
122k1157134
$endgroup$
add a comment |
$begingroup$
$begin{array}\
(1-frac1{k})^{k^2}
&=(frac{k-1}{k})^{k^2}\
&=dfrac1{(frac{k}{k-1})^{k^2}}\
&=dfrac1{(1+frac{1}{k-1})^{k^2}}\
&=dfrac1{((1+frac{1}{k-1})^{k})^k}\
&<dfrac1{(1+frac{k}{k-1})^k}
qquadtext{by Bernoulli}\
&=dfrac1{(frac{2k-1}{k-1})^k}\
&<dfrac1{(frac{2k-2}{k-1})^k}\
&=dfrac1{2^k}\
end{array}
$
and the sum of this converges.
answered 10 mins ago
marty cohenmarty cohen
73.8k549128
$endgroup$
add a comment |
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%2f3123320%2fhow-do-you-determine-if-the-following-series-converges%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT:
Note that $$left( 1-frac1k right)^kle e^{-1}$$
Can you finish?
answered 30 mins ago
Mark ViolaMark Viola
132k1277174
$endgroup$
add a comment |
$begingroup$
HINT:
Note that $$left( 1-frac1k right)^kle e^{-1}$$
Can you finish?
answered 30 mins ago
Mark ViolaMark Viola
132k1277174
$endgroup$
add a comment |
$begingroup$
HINT:
Note that $$left( 1-frac1k right)^kle e^{-1}$$
Can you finish?
answered 30 mins ago
Mark ViolaMark Viola
132k1277174
$endgroup$
HINT:
Note that $$left( 1-frac1k right)^kle e^{-1}$$
Can you finish?
answered 30 mins ago
Mark ViolaMark Viola
132k1277174
answered 30 mins ago
Mark ViolaMark Viola
132k1277174
answered 30 mins ago
Mark ViolaMark Viola
132k1277174
answered 30 mins ago
Mark ViolaMark Viola
132k1277174
132k1277174
add a comment |
add a comment |
$begingroup$
The root test works. Consider
$$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
hence the series converges.
answered 34 mins ago
Theo BenditTheo Bendit
18.7k12253
$endgroup$
add a comment |
$begingroup$
The root test works. Consider
$$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
hence the series converges.
answered 34 mins ago
Theo BenditTheo Bendit
18.7k12253
$endgroup$
add a comment |
$begingroup$
The root test works. Consider
$$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
hence the series converges.
answered 34 mins ago
Theo BenditTheo Bendit
18.7k12253
$endgroup$
The root test works. Consider
$$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
hence the series converges.
answered 34 mins ago
Theo BenditTheo Bendit
18.7k12253
answered 34 mins ago
Theo BenditTheo Bendit
18.7k12253
answered 34 mins ago
Theo BenditTheo Bendit
18.7k12253
answered 34 mins ago
Theo BenditTheo Bendit
18.7k12253
18.7k12253
add a comment |
add a comment |
$begingroup$
The ratio test is also interesting
$$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
$$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$
Develop as a Taylor series for large values of $k$ to get
$$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
Continue with Taylor
$$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$
answered 21 mins ago
Claude LeiboviciClaude Leibovici
122k1157134
$endgroup$
add a comment |
$begingroup$
The ratio test is also interesting
$$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
$$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$
Develop as a Taylor series for large values of $k$ to get
$$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
Continue with Taylor
$$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$
answered 21 mins ago
Claude LeiboviciClaude Leibovici
122k1157134
$endgroup$
add a comment |
$begingroup$
The ratio test is also interesting
$$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
$$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$
Develop as a Taylor series for large values of $k$ to get
$$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
Continue with Taylor
$$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$
answered 21 mins ago
Claude LeiboviciClaude Leibovici
122k1157134
$endgroup$
The ratio test is also interesting
$$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
$$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$
Develop as a Taylor series for large values of $k$ to get
$$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
Continue with Taylor
$$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$
answered 21 mins ago
Claude LeiboviciClaude Leibovici
122k1157134
answered 21 mins ago
Claude LeiboviciClaude Leibovici
122k1157134
answered 21 mins ago
Claude LeiboviciClaude Leibovici
122k1157134
answered 21 mins ago
Claude LeiboviciClaude Leibovici
122k1157134
122k1157134
add a comment |
add a comment |
$begingroup$
$begin{array}\
(1-frac1{k})^{k^2}
&=(frac{k-1}{k})^{k^2}\
&=dfrac1{(frac{k}{k-1})^{k^2}}\
&=dfrac1{(1+frac{1}{k-1})^{k^2}}\
&=dfrac1{((1+frac{1}{k-1})^{k})^k}\
&<dfrac1{(1+frac{k}{k-1})^k}
qquadtext{by Bernoulli}\
&=dfrac1{(frac{2k-1}{k-1})^k}\
&<dfrac1{(frac{2k-2}{k-1})^k}\
&=dfrac1{2^k}\
end{array}
$
and the sum of this converges.
answered 10 mins ago
marty cohenmarty cohen
73.8k549128
$endgroup$
add a comment |
$begingroup$
$begin{array}\
(1-frac1{k})^{k^2}
&=(frac{k-1}{k})^{k^2}\
&=dfrac1{(frac{k}{k-1})^{k^2}}\
&=dfrac1{(1+frac{1}{k-1})^{k^2}}\
&=dfrac1{((1+frac{1}{k-1})^{k})^k}\
&<dfrac1{(1+frac{k}{k-1})^k}
qquadtext{by Bernoulli}\
&=dfrac1{(frac{2k-1}{k-1})^k}\
&<dfrac1{(frac{2k-2}{k-1})^k}\
&=dfrac1{2^k}\
end{array}
$
and the sum of this converges.
answered 10 mins ago
marty cohenmarty cohen
73.8k549128
$endgroup$
add a comment |
$begingroup$
$begin{array}\
(1-frac1{k})^{k^2}
&=(frac{k-1}{k})^{k^2}\
&=dfrac1{(frac{k}{k-1})^{k^2}}\
&=dfrac1{(1+frac{1}{k-1})^{k^2}}\
&=dfrac1{((1+frac{1}{k-1})^{k})^k}\
&<dfrac1{(1+frac{k}{k-1})^k}
qquadtext{by Bernoulli}\
&=dfrac1{(frac{2k-1}{k-1})^k}\
&<dfrac1{(frac{2k-2}{k-1})^k}\
&=dfrac1{2^k}\
end{array}
$
and the sum of this converges.
answered 10 mins ago
marty cohenmarty cohen
73.8k549128
$endgroup$
$begin{array}\
(1-frac1{k})^{k^2}
&=(frac{k-1}{k})^{k^2}\
&=dfrac1{(frac{k}{k-1})^{k^2}}\
&=dfrac1{(1+frac{1}{k-1})^{k^2}}\
&=dfrac1{((1+frac{1}{k-1})^{k})^k}\
&<dfrac1{(1+frac{k}{k-1})^k}
qquadtext{by Bernoulli}\
&=dfrac1{(frac{2k-1}{k-1})^k}\
&<dfrac1{(frac{2k-2}{k-1})^k}\
&=dfrac1{2^k}\
end{array}
$
and the sum of this converges.
answered 10 mins ago
marty cohenmarty cohen
73.8k549128
answered 10 mins ago
marty cohenmarty cohen
73.8k549128
answered 10 mins ago
marty cohenmarty cohen
73.8k549128
answered 10 mins ago
marty cohenmarty cohen
73.8k549128
73.8k549128
add a comment |
add a comment |
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.
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%2f3123320%2fhow-do-you-determine-if-the-following-series-converges%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
$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
35 mins ago
$begingroup$
Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
$endgroup$
– JavaMan
29 mins ago