Alternative proof for $ sum left(frac{k-2}{k}right)^k$ is divergent.
$begingroup$
In an earlier question I asked help to prove that:
$$ sum left(frac{k-2}{k}right)^k$$ is divergent, using the fact the the general term converges to $frac{1}{e^2}$ and by the limit test the series diverges, is there a more straightforward method I am missing? One that does not use $e$ necessarily.
See original post: Series diverges
real-analysis sequences-and-series alternative-proof
$endgroup$
add a comment |
$begingroup$
In an earlier question I asked help to prove that:
$$ sum left(frac{k-2}{k}right)^k$$ is divergent, using the fact the the general term converges to $frac{1}{e^2}$ and by the limit test the series diverges, is there a more straightforward method I am missing? One that does not use $e$ necessarily.
See original post: Series diverges
real-analysis sequences-and-series alternative-proof
$endgroup$
10
$begingroup$
I would say that proving that the general term does not converge to zero is about as simple as it gets.
$endgroup$
– MisterRiemann
Nov 28 '18 at 12:17
add a comment |
$begingroup$
In an earlier question I asked help to prove that:
$$ sum left(frac{k-2}{k}right)^k$$ is divergent, using the fact the the general term converges to $frac{1}{e^2}$ and by the limit test the series diverges, is there a more straightforward method I am missing? One that does not use $e$ necessarily.
See original post: Series diverges
real-analysis sequences-and-series alternative-proof
$endgroup$
In an earlier question I asked help to prove that:
$$ sum left(frac{k-2}{k}right)^k$$ is divergent, using the fact the the general term converges to $frac{1}{e^2}$ and by the limit test the series diverges, is there a more straightforward method I am missing? One that does not use $e$ necessarily.
See original post: Series diverges
real-analysis sequences-and-series alternative-proof
real-analysis sequences-and-series alternative-proof
asked Nov 28 '18 at 12:11
Wesley StrikWesley Strik
1,635423
1,635423
10
$begingroup$
I would say that proving that the general term does not converge to zero is about as simple as it gets.
$endgroup$
– MisterRiemann
Nov 28 '18 at 12:17
add a comment |
10
$begingroup$
I would say that proving that the general term does not converge to zero is about as simple as it gets.
$endgroup$
– MisterRiemann
Nov 28 '18 at 12:17
10
10
$begingroup$
I would say that proving that the general term does not converge to zero is about as simple as it gets.
$endgroup$
– MisterRiemann
Nov 28 '18 at 12:17
$begingroup$
I would say that proving that the general term does not converge to zero is about as simple as it gets.
$endgroup$
– MisterRiemann
Nov 28 '18 at 12:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
if $n>2$ then $Big(dfrac{n-2}{n}Big)^n>0$ and
$$
\lim_{nto+infty}{Big(frac{n-2}{n}Big)^n}=lim_{nto+infty}{Big(1-frac{2}{n}Big)^n}=e^{-2}>0
$$
$endgroup$
$begingroup$
This is the approach I used in that question, so it does not really answer my question.
$endgroup$
– Wesley Strik
Nov 28 '18 at 20:06
$begingroup$
Then why did you accept it?
$endgroup$
– marty cohen
Nov 28 '18 at 21:18
add a comment |
$begingroup$
Suppose we don't know anything about $e$. We can still show that $(1-2/k)^knotto0$ as $ktoinfty$, using the fact that the (alternating) binomial expansion can be truncated after a few terms to give an inequality, since $2^k/k!lt1$ for $kge4$.
$$begin{align}
left(k-2over k right)^k
&=left(1-{2over k}right)^k\
&=1-{kchoose1}left(2over kright)+{kchoose2}left(2over kright)^2-{kchoose3}left(2over kright)^3+cdots\
&=1-2+2-{8over6}{(k-1)(k-2)over k^2}+{16over24}{(k-1)(k-2)(k-3)over k^3}-{32over120}{(k-1)(k-2)(k-3)(k-4)over k^4}+cdots\
>1-2+2-{4over3}+{2over3}left(1-{6over k}right)-{4over15}\
&={1over15}-{4over k}
end{align}$$
A nice way to wrap things up is now to note that, for $kge105$, we have
$$left(k-2over kright)^kgt{1over15}-{4over105}={1over35}$$
and therefore $(1-2/k)^knotto0$ as $ktoinfty$, which in turn implies $sum(1-2/k)^k$ diverges.
Remarks: In addition to $2^k/k!lt1$ for $kge4$, which permitted truncation after a subtraction, we used the crude inequalities $(k-1)(k-2)lt k^2$ and $(k-1)(k-2)(k-3)(k-4)lt k^4$ and the slightly less crude inequality $(k-1)(k-2)(k-3)=k^3-6k^2+11k-6gt k^3-6k^2$.
$endgroup$
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%2f3017071%2falternative-proof-for-sum-left-frack-2k-rightk-is-divergent%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
if $n>2$ then $Big(dfrac{n-2}{n}Big)^n>0$ and
$$
\lim_{nto+infty}{Big(frac{n-2}{n}Big)^n}=lim_{nto+infty}{Big(1-frac{2}{n}Big)^n}=e^{-2}>0
$$
$endgroup$
$begingroup$
This is the approach I used in that question, so it does not really answer my question.
$endgroup$
– Wesley Strik
Nov 28 '18 at 20:06
$begingroup$
Then why did you accept it?
$endgroup$
– marty cohen
Nov 28 '18 at 21:18
add a comment |
$begingroup$
if $n>2$ then $Big(dfrac{n-2}{n}Big)^n>0$ and
$$
\lim_{nto+infty}{Big(frac{n-2}{n}Big)^n}=lim_{nto+infty}{Big(1-frac{2}{n}Big)^n}=e^{-2}>0
$$
$endgroup$
$begingroup$
This is the approach I used in that question, so it does not really answer my question.
$endgroup$
– Wesley Strik
Nov 28 '18 at 20:06
$begingroup$
Then why did you accept it?
$endgroup$
– marty cohen
Nov 28 '18 at 21:18
add a comment |
$begingroup$
if $n>2$ then $Big(dfrac{n-2}{n}Big)^n>0$ and
$$
\lim_{nto+infty}{Big(frac{n-2}{n}Big)^n}=lim_{nto+infty}{Big(1-frac{2}{n}Big)^n}=e^{-2}>0
$$
$endgroup$
if $n>2$ then $Big(dfrac{n-2}{n}Big)^n>0$ and
$$
\lim_{nto+infty}{Big(frac{n-2}{n}Big)^n}=lim_{nto+infty}{Big(1-frac{2}{n}Big)^n}=e^{-2}>0
$$
answered Nov 28 '18 at 20:01
Samvel SafaryanSamvel Safaryan
501111
501111
$begingroup$
This is the approach I used in that question, so it does not really answer my question.
$endgroup$
– Wesley Strik
Nov 28 '18 at 20:06
$begingroup$
Then why did you accept it?
$endgroup$
– marty cohen
Nov 28 '18 at 21:18
add a comment |
$begingroup$
This is the approach I used in that question, so it does not really answer my question.
$endgroup$
– Wesley Strik
Nov 28 '18 at 20:06
$begingroup$
Then why did you accept it?
$endgroup$
– marty cohen
Nov 28 '18 at 21:18
$begingroup$
This is the approach I used in that question, so it does not really answer my question.
$endgroup$
– Wesley Strik
Nov 28 '18 at 20:06
$begingroup$
This is the approach I used in that question, so it does not really answer my question.
$endgroup$
– Wesley Strik
Nov 28 '18 at 20:06
$begingroup$
Then why did you accept it?
$endgroup$
– marty cohen
Nov 28 '18 at 21:18
$begingroup$
Then why did you accept it?
$endgroup$
– marty cohen
Nov 28 '18 at 21:18
add a comment |
$begingroup$
Suppose we don't know anything about $e$. We can still show that $(1-2/k)^knotto0$ as $ktoinfty$, using the fact that the (alternating) binomial expansion can be truncated after a few terms to give an inequality, since $2^k/k!lt1$ for $kge4$.
$$begin{align}
left(k-2over k right)^k
&=left(1-{2over k}right)^k\
&=1-{kchoose1}left(2over kright)+{kchoose2}left(2over kright)^2-{kchoose3}left(2over kright)^3+cdots\
&=1-2+2-{8over6}{(k-1)(k-2)over k^2}+{16over24}{(k-1)(k-2)(k-3)over k^3}-{32over120}{(k-1)(k-2)(k-3)(k-4)over k^4}+cdots\
>1-2+2-{4over3}+{2over3}left(1-{6over k}right)-{4over15}\
&={1over15}-{4over k}
end{align}$$
A nice way to wrap things up is now to note that, for $kge105$, we have
$$left(k-2over kright)^kgt{1over15}-{4over105}={1over35}$$
and therefore $(1-2/k)^knotto0$ as $ktoinfty$, which in turn implies $sum(1-2/k)^k$ diverges.
Remarks: In addition to $2^k/k!lt1$ for $kge4$, which permitted truncation after a subtraction, we used the crude inequalities $(k-1)(k-2)lt k^2$ and $(k-1)(k-2)(k-3)(k-4)lt k^4$ and the slightly less crude inequality $(k-1)(k-2)(k-3)=k^3-6k^2+11k-6gt k^3-6k^2$.
$endgroup$
add a comment |
$begingroup$
Suppose we don't know anything about $e$. We can still show that $(1-2/k)^knotto0$ as $ktoinfty$, using the fact that the (alternating) binomial expansion can be truncated after a few terms to give an inequality, since $2^k/k!lt1$ for $kge4$.
$$begin{align}
left(k-2over k right)^k
&=left(1-{2over k}right)^k\
&=1-{kchoose1}left(2over kright)+{kchoose2}left(2over kright)^2-{kchoose3}left(2over kright)^3+cdots\
&=1-2+2-{8over6}{(k-1)(k-2)over k^2}+{16over24}{(k-1)(k-2)(k-3)over k^3}-{32over120}{(k-1)(k-2)(k-3)(k-4)over k^4}+cdots\
>1-2+2-{4over3}+{2over3}left(1-{6over k}right)-{4over15}\
&={1over15}-{4over k}
end{align}$$
A nice way to wrap things up is now to note that, for $kge105$, we have
$$left(k-2over kright)^kgt{1over15}-{4over105}={1over35}$$
and therefore $(1-2/k)^knotto0$ as $ktoinfty$, which in turn implies $sum(1-2/k)^k$ diverges.
Remarks: In addition to $2^k/k!lt1$ for $kge4$, which permitted truncation after a subtraction, we used the crude inequalities $(k-1)(k-2)lt k^2$ and $(k-1)(k-2)(k-3)(k-4)lt k^4$ and the slightly less crude inequality $(k-1)(k-2)(k-3)=k^3-6k^2+11k-6gt k^3-6k^2$.
$endgroup$
add a comment |
$begingroup$
Suppose we don't know anything about $e$. We can still show that $(1-2/k)^knotto0$ as $ktoinfty$, using the fact that the (alternating) binomial expansion can be truncated after a few terms to give an inequality, since $2^k/k!lt1$ for $kge4$.
$$begin{align}
left(k-2over k right)^k
&=left(1-{2over k}right)^k\
&=1-{kchoose1}left(2over kright)+{kchoose2}left(2over kright)^2-{kchoose3}left(2over kright)^3+cdots\
&=1-2+2-{8over6}{(k-1)(k-2)over k^2}+{16over24}{(k-1)(k-2)(k-3)over k^3}-{32over120}{(k-1)(k-2)(k-3)(k-4)over k^4}+cdots\
>1-2+2-{4over3}+{2over3}left(1-{6over k}right)-{4over15}\
&={1over15}-{4over k}
end{align}$$
A nice way to wrap things up is now to note that, for $kge105$, we have
$$left(k-2over kright)^kgt{1over15}-{4over105}={1over35}$$
and therefore $(1-2/k)^knotto0$ as $ktoinfty$, which in turn implies $sum(1-2/k)^k$ diverges.
Remarks: In addition to $2^k/k!lt1$ for $kge4$, which permitted truncation after a subtraction, we used the crude inequalities $(k-1)(k-2)lt k^2$ and $(k-1)(k-2)(k-3)(k-4)lt k^4$ and the slightly less crude inequality $(k-1)(k-2)(k-3)=k^3-6k^2+11k-6gt k^3-6k^2$.
$endgroup$
Suppose we don't know anything about $e$. We can still show that $(1-2/k)^knotto0$ as $ktoinfty$, using the fact that the (alternating) binomial expansion can be truncated after a few terms to give an inequality, since $2^k/k!lt1$ for $kge4$.
$$begin{align}
left(k-2over k right)^k
&=left(1-{2over k}right)^k\
&=1-{kchoose1}left(2over kright)+{kchoose2}left(2over kright)^2-{kchoose3}left(2over kright)^3+cdots\
&=1-2+2-{8over6}{(k-1)(k-2)over k^2}+{16over24}{(k-1)(k-2)(k-3)over k^3}-{32over120}{(k-1)(k-2)(k-3)(k-4)over k^4}+cdots\
>1-2+2-{4over3}+{2over3}left(1-{6over k}right)-{4over15}\
&={1over15}-{4over k}
end{align}$$
A nice way to wrap things up is now to note that, for $kge105$, we have
$$left(k-2over kright)^kgt{1over15}-{4over105}={1over35}$$
and therefore $(1-2/k)^knotto0$ as $ktoinfty$, which in turn implies $sum(1-2/k)^k$ diverges.
Remarks: In addition to $2^k/k!lt1$ for $kge4$, which permitted truncation after a subtraction, we used the crude inequalities $(k-1)(k-2)lt k^2$ and $(k-1)(k-2)(k-3)(k-4)lt k^4$ and the slightly less crude inequality $(k-1)(k-2)(k-3)=k^3-6k^2+11k-6gt k^3-6k^2$.
answered Nov 28 '18 at 22:46
Barry CipraBarry Cipra
59.3k653125
59.3k653125
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%2f3017071%2falternative-proof-for-sum-left-frack-2k-rightk-is-divergent%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
10
$begingroup$
I would say that proving that the general term does not converge to zero is about as simple as it gets.
$endgroup$
– MisterRiemann
Nov 28 '18 at 12:17