Dtjytyk

Was Stan Lee's cameo in Captain Marvel computer generated?

Difference between `nmap local-IP-address` and `nmap localhost`

What would be the most expensive material to an intergalactic society?

If sound is a longitudinal wave, why can we hear it if our ears aren't aligned with the propagation direction?

What does *dead* mean in *What do you mean, dead?*?

What can I do if someone tampers with my SSH public key?

Psychology of 'flow'

Computation logic of Partway in TikZ

Professor forcing me to attend a conference, I can't afford even with 50% funding

Is there a math expression equivalent to the conditional ternary operator?

Locked Away- What am I?

Was it really inappropriate to write a pull request for the company I interviewed with?

Is "cogitate" used appropriately in "I cogitate that success relies on hard work"?

Is there a logarithm base for which the logarithm becomes an identity function?

Called into a meeting and told we are being made redundant (laid off) and "not to share outside". Can I tell my partner?

Either of .... (Plural/Singular)

Prove this function has at most two zero points

Should I apply for my boss's promotion?

A running toilet that stops itself

Is it a Cyclops number? "Nobody" knows!

Rationale to prefer local variables over instance variables?

I've given my players a lot of magic items. Is it reasonable for me to give them harder encounters?

Why do phishing e-mails use faked e-mail addresses instead of the real one?

Two swapfiles, hibernate working, how to correctly configure hibernate to not run out of space, or fail somehow

Proving convergence/divergence of this series?

hard time with series convergence or divergenceSeries convergence/divergenceProving uniform convergence of this seriesProving convergence/divergence for $p$-seriesProving the convergence/divergence of a seemingly oscillating seriesFinding the convergence/divergence of a seriesConvergence of series/absolute convergenceShow the convergence or divergence of seriesDivergent infinite series $n!e^n/n^n$ - simpler proof of divergence?Help with convergence tests for series

3

3

$begingroup$

$$sum_{k=0}^{infty}frac{(k!)}{(k+1)!+2^k}$$

Anyone know how to show that this series converges or diverges? I've tried multiple tests and nothing seems to be working. Any help would be amazing!

sequences-and-series convergence

share|cite|improve this question

edited yesterday

mrtaurho

5,74551540

asked yesterday

Bob JonesBob Jones

192

New contributor

Bob Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

3

3

$begingroup$

$$sum_{k=0}^{infty}frac{(k!)}{(k+1)!+2^k}$$

Anyone know how to show that this series converges or diverges? I've tried multiple tests and nothing seems to be working. Any help would be amazing!

sequences-and-series convergence

share|cite|improve this question

edited yesterday

mrtaurho

5,74551540

asked yesterday

Bob JonesBob Jones

192

New contributor

Bob Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

$$sum_{k=0}^{infty}frac{(k!)}{(k+1)!+2^k}$$

Anyone know how to show that this series converges or diverges? I've tried multiple tests and nothing seems to be working. Any help would be amazing!

sequences-and-series convergence

share|cite|improve this question

edited yesterday

mrtaurho

5,74551540

asked yesterday

Bob JonesBob Jones

192

New contributor

Bob Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited yesterday

mrtaurho

5,74551540

asked yesterday

Bob JonesBob Jones

192

New contributor

Bob Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited yesterday

mrtaurho

5,74551540

asked yesterday

Bob JonesBob Jones

192

New contributor

Bob Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited yesterday

mrtaurho

5,74551540

edited yesterday

mrtaurho

5,74551540

asked yesterday

Bob JonesBob Jones

192

New contributor

Bob Jones is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

Bob JonesBob Jones

192

2 Answers
2

active

oldest

votes

1

$begingroup$

Since $(k+1)!$ grows faster than $2^k$ there exist some $K_0$ so that the denominator is equivalent to $(k+1)!$ for $forall kgeq K_0$ then:$sum_0^inftyfrac{k!}{(k+1)!+2^k} simsum_{K_0}^infty frac{1}{k+1}rightarrowinfty$

Edit: I'll clarify my steps as suggested.

$(1)$ $k!ge 2^k$ for big enough k

Proof:

Consider $f_k=dfrac{k!}{2^k}$; the Limit $displaystyle{lim_{krightarrowinfty}dfrac{f_{k+1}}{f_k}=lim_{ktoinfty}dfrac{k+1}{2}=infty};$

By the ratio test $displaystyle{lim_{ktoinfty}f_k=infty}$

So by the precise definition of a limit there exist some $K_0inmathbb{N}$ so that for all $kge K_0Rightarrow k!ge2^k$

We also get that $displaystyle{lim_{ktoinfty}frac{1}{f_k}=lim_{ktoinfty}frac{2^k}{k!}=0}$

(2) "Let $a_k=dfrac{k!}{(k+1)!+2^k}$and $b_k=dfrac{k!}{(k+1)!}$ then if $displaystyle{lim_{ktoinfty}frac{a_k}{b_k}=L, Lin(0,infty)}$ by the limit comparision test the series associated to $a_k,b_k$ either both diverge or both converge."

Proof that the limit exist:

$displaystyle{lim_{ktoinfty}frac{a_k}{b_k}=lim_{ktoinfty}frac{(k+1)!}{(k+1)!+2^k}=lim_{ktoinfty}cfrac{1}{1+dfrac{2^k}{(k+1)!}}=dfrac{1}{displaystyle{lim_{ktoinfty}1+frac{2^k}{(k+1)!}}}}$.

$displaystyle{lim_{ktoinfty}frac{2^k}{(k+1)!}}=lim_{ktoinfty}frac{1}{k+1}frac{2^k}{k!}=0cdot0=0$

We get that $displaystyle{lim_{ktoinfty}frac{a_k}{b_k}=1}$

So by the limit comparision test:
$$sum_{K_0}^{infty}b_k=sum_{K_0}^{infty}dfrac{1}{k+1}=inftyRightarrow sum_{K_0}^{infty}a_k=infty$$

share|cite|improve this answer

edited 2 hours ago

BarzanHayati

1458

answered yesterday

Sebastian CorSebastian Cor

948

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please edit denominator of fraction in series. I think you must explain more about this approximation
  $endgroup$
  – BarzanHayati
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $sum_0^inftyfrac{k!}{(k+1)!+2^k} < sum_{0}^infty frac{1}{k+1}rightarrow$. But how to conclude that it would be a divergent series?
  $endgroup$
  – BarzanHayati
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I edited my answer, I hope it is clear now.
  $endgroup$
  – Sebastian Cor
  yesterday
  

  
  
  
  

add a comment | 

0

$begingroup$

I suggest to divide the numerator and the denominator of the fraction by $k!$:
$$
frac{k!}{(k+1)! + 2^k} = frac{1}{k+1 + frac{2^k}{k!}} geq frac{1}{k+3} ,
$$
since $frac{2^k}{k!} leq 2$ for all positive integer $k$.

Thus
$$
sum_{k=0}^{infty} frac{k!}{(k+1)! + 2^k} geq sum_{k=0}^{infty} frac{1}{k+3} .
$$

share|cite|improve this answer

answered 23 hours ago

ErtxiemErtxiem

585

$endgroup$

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
});


}
});

Bob Jones is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3139620%2fproving-convergence-divergence-of-this-series%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

Since $(k+1)!$ grows faster than $2^k$ there exist some $K_0$ so that the denominator is equivalent to $(k+1)!$ for $forall kgeq K_0$ then:$sum_0^inftyfrac{k!}{(k+1)!+2^k} simsum_{K_0}^infty frac{1}{k+1}rightarrowinfty$

Edit: I'll clarify my steps as suggested.

$(1)$ $k!ge 2^k$ for big enough k

Proof:

Consider $f_k=dfrac{k!}{2^k}$; the Limit $displaystyle{lim_{krightarrowinfty}dfrac{f_{k+1}}{f_k}=lim_{ktoinfty}dfrac{k+1}{2}=infty};$

By the ratio test $displaystyle{lim_{ktoinfty}f_k=infty}$

So by the precise definition of a limit there exist some $K_0inmathbb{N}$ so that for all $kge K_0Rightarrow k!ge2^k$

We also get that $displaystyle{lim_{ktoinfty}frac{1}{f_k}=lim_{ktoinfty}frac{2^k}{k!}=0}$

(2) "Let $a_k=dfrac{k!}{(k+1)!+2^k}$and $b_k=dfrac{k!}{(k+1)!}$ then if $displaystyle{lim_{ktoinfty}frac{a_k}{b_k}=L, Lin(0,infty)}$ by the limit comparision test the series associated to $a_k,b_k$ either both diverge or both converge."

Proof that the limit exist:

$displaystyle{lim_{ktoinfty}frac{a_k}{b_k}=lim_{ktoinfty}frac{(k+1)!}{(k+1)!+2^k}=lim_{ktoinfty}cfrac{1}{1+dfrac{2^k}{(k+1)!}}=dfrac{1}{displaystyle{lim_{ktoinfty}1+frac{2^k}{(k+1)!}}}}$.

$displaystyle{lim_{ktoinfty}frac{2^k}{(k+1)!}}=lim_{ktoinfty}frac{1}{k+1}frac{2^k}{k!}=0cdot0=0$

We get that $displaystyle{lim_{ktoinfty}frac{a_k}{b_k}=1}$

So by the limit comparision test:
$$sum_{K_0}^{infty}b_k=sum_{K_0}^{infty}dfrac{1}{k+1}=inftyRightarrow sum_{K_0}^{infty}a_k=infty$$

share|cite|improve this answer

edited 2 hours ago

BarzanHayati

1458

answered yesterday

Sebastian CorSebastian Cor

948

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please edit denominator of fraction in series. I think you must explain more about this approximation
  $endgroup$
  – BarzanHayati
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $sum_0^inftyfrac{k!}{(k+1)!+2^k} < sum_{0}^infty frac{1}{k+1}rightarrow$. But how to conclude that it would be a divergent series?
  $endgroup$
  – BarzanHayati
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I edited my answer, I hope it is clear now.
  $endgroup$
  – Sebastian Cor
  yesterday
  

  
  
  
  

add a comment | 

1

$begingroup$

Since $(k+1)!$ grows faster than $2^k$ there exist some $K_0$ so that the denominator is equivalent to $(k+1)!$ for $forall kgeq K_0$ then:$sum_0^inftyfrac{k!}{(k+1)!+2^k} simsum_{K_0}^infty frac{1}{k+1}rightarrowinfty$

Edit: I'll clarify my steps as suggested.

$(1)$ $k!ge 2^k$ for big enough k

Proof:

Consider $f_k=dfrac{k!}{2^k}$; the Limit $displaystyle{lim_{krightarrowinfty}dfrac{f_{k+1}}{f_k}=lim_{ktoinfty}dfrac{k+1}{2}=infty};$

By the ratio test $displaystyle{lim_{ktoinfty}f_k=infty}$

So by the precise definition of a limit there exist some $K_0inmathbb{N}$ so that for all $kge K_0Rightarrow k!ge2^k$

We also get that $displaystyle{lim_{ktoinfty}frac{1}{f_k}=lim_{ktoinfty}frac{2^k}{k!}=0}$

(2) "Let $a_k=dfrac{k!}{(k+1)!+2^k}$and $b_k=dfrac{k!}{(k+1)!}$ then if $displaystyle{lim_{ktoinfty}frac{a_k}{b_k}=L, Lin(0,infty)}$ by the limit comparision test the series associated to $a_k,b_k$ either both diverge or both converge."

Proof that the limit exist:

$displaystyle{lim_{ktoinfty}frac{a_k}{b_k}=lim_{ktoinfty}frac{(k+1)!}{(k+1)!+2^k}=lim_{ktoinfty}cfrac{1}{1+dfrac{2^k}{(k+1)!}}=dfrac{1}{displaystyle{lim_{ktoinfty}1+frac{2^k}{(k+1)!}}}}$.

$displaystyle{lim_{ktoinfty}frac{2^k}{(k+1)!}}=lim_{ktoinfty}frac{1}{k+1}frac{2^k}{k!}=0cdot0=0$

We get that $displaystyle{lim_{ktoinfty}frac{a_k}{b_k}=1}$

So by the limit comparision test:
$$sum_{K_0}^{infty}b_k=sum_{K_0}^{infty}dfrac{1}{k+1}=inftyRightarrow sum_{K_0}^{infty}a_k=infty$$

share|cite|improve this answer

edited 2 hours ago

BarzanHayati

1458

answered yesterday

Sebastian CorSebastian Cor

948

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please edit denominator of fraction in series. I think you must explain more about this approximation
  $endgroup$
  – BarzanHayati
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $sum_0^inftyfrac{k!}{(k+1)!+2^k} < sum_{0}^infty frac{1}{k+1}rightarrow$. But how to conclude that it would be a divergent series?
  $endgroup$
  – BarzanHayati
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I edited my answer, I hope it is clear now.
  $endgroup$
  – Sebastian Cor
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Since $(k+1)!$ grows faster than $2^k$ there exist some $K_0$ so that the denominator is equivalent to $(k+1)!$ for $forall kgeq K_0$ then:$sum_0^inftyfrac{k!}{(k+1)!+2^k} simsum_{K_0}^infty frac{1}{k+1}rightarrowinfty$

Edit: I'll clarify my steps as suggested.

$(1)$ $k!ge 2^k$ for big enough k

Proof:

Consider $f_k=dfrac{k!}{2^k}$; the Limit $displaystyle{lim_{krightarrowinfty}dfrac{f_{k+1}}{f_k}=lim_{ktoinfty}dfrac{k+1}{2}=infty};$

By the ratio test $displaystyle{lim_{ktoinfty}f_k=infty}$

So by the precise definition of a limit there exist some $K_0inmathbb{N}$ so that for all $kge K_0Rightarrow k!ge2^k$

We also get that $displaystyle{lim_{ktoinfty}frac{1}{f_k}=lim_{ktoinfty}frac{2^k}{k!}=0}$

(2) "Let $a_k=dfrac{k!}{(k+1)!+2^k}$and $b_k=dfrac{k!}{(k+1)!}$ then if $displaystyle{lim_{ktoinfty}frac{a_k}{b_k}=L, Lin(0,infty)}$ by the limit comparision test the series associated to $a_k,b_k$ either both diverge or both converge."

Proof that the limit exist:

$displaystyle{lim_{ktoinfty}frac{a_k}{b_k}=lim_{ktoinfty}frac{(k+1)!}{(k+1)!+2^k}=lim_{ktoinfty}cfrac{1}{1+dfrac{2^k}{(k+1)!}}=dfrac{1}{displaystyle{lim_{ktoinfty}1+frac{2^k}{(k+1)!}}}}$.

$displaystyle{lim_{ktoinfty}frac{2^k}{(k+1)!}}=lim_{ktoinfty}frac{1}{k+1}frac{2^k}{k!}=0cdot0=0$

We get that $displaystyle{lim_{ktoinfty}frac{a_k}{b_k}=1}$

So by the limit comparision test:
$$sum_{K_0}^{infty}b_k=sum_{K_0}^{infty}dfrac{1}{k+1}=inftyRightarrow sum_{K_0}^{infty}a_k=infty$$

share|cite|improve this answer

edited 2 hours ago

BarzanHayati

1458

answered yesterday

Sebastian CorSebastian Cor

948

$endgroup$

share|cite|improve this answer

edited 2 hours ago

BarzanHayati

1458

answered yesterday

Sebastian CorSebastian Cor

948

edited 2 hours ago

BarzanHayati

1458

edited 2 hours ago

BarzanHayati

1458

answered yesterday

Sebastian CorSebastian Cor

948

answered yesterday

Sebastian CorSebastian Cor

948

0

$begingroup$

I suggest to divide the numerator and the denominator of the fraction by $k!$:
$$
frac{k!}{(k+1)! + 2^k} = frac{1}{k+1 + frac{2^k}{k!}} geq frac{1}{k+3} ,
$$
since $frac{2^k}{k!} leq 2$ for all positive integer $k$.

Thus
$$
sum_{k=0}^{infty} frac{k!}{(k+1)! + 2^k} geq sum_{k=0}^{infty} frac{1}{k+3} .
$$

share|cite|improve this answer

answered 23 hours ago

ErtxiemErtxiem

585

$endgroup$

add a comment | 

0

$begingroup$

I suggest to divide the numerator and the denominator of the fraction by $k!$:
$$
frac{k!}{(k+1)! + 2^k} = frac{1}{k+1 + frac{2^k}{k!}} geq frac{1}{k+3} ,
$$
since $frac{2^k}{k!} leq 2$ for all positive integer $k$.

Thus
$$
sum_{k=0}^{infty} frac{k!}{(k+1)! + 2^k} geq sum_{k=0}^{infty} frac{1}{k+3} .
$$

share|cite|improve this answer

answered 23 hours ago

ErtxiemErtxiem

585

$endgroup$

add a comment | 

$begingroup$

I suggest to divide the numerator and the denominator of the fraction by $k!$:
$$
frac{k!}{(k+1)! + 2^k} = frac{1}{k+1 + frac{2^k}{k!}} geq frac{1}{k+3} ,
$$
since $frac{2^k}{k!} leq 2$ for all positive integer $k$.

Thus
$$
sum_{k=0}^{infty} frac{k!}{(k+1)! + 2^k} geq sum_{k=0}^{infty} frac{1}{k+3} .
$$

share|cite|improve this answer

answered 23 hours ago

ErtxiemErtxiem

585

$endgroup$

share|cite|improve this answer

answered 23 hours ago

ErtxiemErtxiem

585

answered 23 hours ago

ErtxiemErtxiem

585

answered 23 hours ago

ErtxiemErtxiem

585