Hfrhyu

Is it tax fraud for an individual to declare non-taxable revenue as taxable income? (US tax laws)

How old can references or sources in a thesis be?

Why doesn't Newton's third law mean a person bounces back to where they started when they hit the ground?

declaring a variable twice in IIFE

How is the claim "I am in New York only if I am in America" the same as "If I am in New York, then I am in America?

How did the USSR manage to innovate in an environment characterized by government censorship and high bureaucracy?

GPS Rollover on Android Smartphones

Why linear maps act like matrix multiplication?

Proof for divisibility of polynomials.

N.B. ligature in Latex

Is there any sparring that doesn't involve punches to the head?

Why is the design of haulage companies so “special”?

Why don't electromagnetic waves interact with each other?

Explain the parameters before and after @ in the terminal prompt

Mathematical cryptic clues

What is the offset in a seaplane's hull?

How can I hide my bitcoin transactions to protect anonymity from others?

How can bays and straits be determined in a procedurally generated map?

How to get the available space of $HOME as a variable in shell scripting?

Possibly bubble sort algorithm

Do any Labour MPs support no-deal?

Can I make popcorn with any corn?

Is it possible to do 50 km distance without any previous training?

Infinite past with a beginning?

how to write the general formula for a telescoping series

Any idea how to determine the convergence of the following series?How do I calculate the value of this series?Telescoping series order.Solving Telescoping SeriesLet $a_n to 0$. Prove that the series $sum_n=0^infty a_n x^n$ converges uniformly on $|x|le 1/2$For which $n$ the given series convergesEvaluate $sum_k=1^infty frack-12^k+1$ as a telescoping series?Help summing the telescoping series $sum_n=2^inftyfrac1n^3-n$.Help with convergence tests for seriesThe Convergence of a Telescoping Series

1

$begingroup$

Im trying to check if the series

$sum_k=2^infty frac1sqrtk-1 - frac1sqrtk+1 $

is converging or not.

I have divided the series into two parts with the series with even k >= 2 and the series with the odd k >= 3.

For the series with even k >=2 I have started writing down the terms:

$S_k = (1-frac1sqrt3)+ (frac1sqrt3-frac1sqrt5)+(frac1sqrt5-frac1sqrt7)+...$

But I am wodering how I can write down the last two terms (k-1, and k) to show that this series converges.

sequences-and-series convergence summation telescopic-series

share|cite|improve this question

edited Apr 3 at 9:42

José Carlos Santos

173k23133241

asked Apr 3 at 8:54

F WiF Wi

474

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Try to maybe write out the first couple of cases: $S_1$, $S_2$ --- maybe $S_5$ .. !
  $endgroup$
  – Matti P.
  Apr 3 at 8:57
  

  
  
  
  

add a comment | 

1

$begingroup$

Im trying to check if the series

$sum_k=2^infty frac1sqrtk-1 - frac1sqrtk+1 $

is converging or not.

I have divided the series into two parts with the series with even k >= 2 and the series with the odd k >= 3.

For the series with even k >=2 I have started writing down the terms:

$S_k = (1-frac1sqrt3)+ (frac1sqrt3-frac1sqrt5)+(frac1sqrt5-frac1sqrt7)+...$

But I am wodering how I can write down the last two terms (k-1, and k) to show that this series converges.

sequences-and-series convergence summation telescopic-series

share|cite|improve this question

edited Apr 3 at 9:42

José Carlos Santos

173k23133241

asked Apr 3 at 8:54

F WiF Wi

474

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Try to maybe write out the first couple of cases: $S_1$, $S_2$ --- maybe $S_5$ .. !
  $endgroup$
  – Matti P.
  Apr 3 at 8:57
  

  
  
  
  

add a comment | 

$begingroup$

Im trying to check if the series

$sum_k=2^infty frac1sqrtk-1 - frac1sqrtk+1 $

is converging or not.

I have divided the series into two parts with the series with even k >= 2 and the series with the odd k >= 3.

For the series with even k >=2 I have started writing down the terms:

$S_k = (1-frac1sqrt3)+ (frac1sqrt3-frac1sqrt5)+(frac1sqrt5-frac1sqrt7)+...$

But I am wodering how I can write down the last two terms (k-1, and k) to show that this series converges.

sequences-and-series convergence summation telescopic-series

share|cite|improve this question

edited Apr 3 at 9:42

José Carlos Santos

173k23133241

asked Apr 3 at 8:54

F WiF Wi

474

$endgroup$

share|cite|improve this question

edited Apr 3 at 9:42

José Carlos Santos

173k23133241

asked Apr 3 at 8:54

F WiF Wi

474

share|cite|improve this question

edited Apr 3 at 9:42

José Carlos Santos

173k23133241

asked Apr 3 at 8:54

F WiF Wi

474

edited Apr 3 at 9:42

José Carlos Santos

173k23133241

edited Apr 3 at 9:42

José Carlos Santos

173k23133241

asked Apr 3 at 8:54

F WiF Wi

474

asked Apr 3 at 8:54

F WiF Wi

474

2 Answers
2

active

oldest

votes

2

$begingroup$

Since$$sum_k=2^inftyleft(frac1sqrtk-1-frac1sqrtk+1right)=sum_k=2^inftyleft(frac1sqrtk-1-frac1sqrt kright)+sum_k=2^inftyleft(frac1sqrt k-frac1sqrtk+1right),$$the sum of your series is $1+dfrac1sqrt2$.

share|cite|improve this answer

answered Apr 3 at 8:57

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thgis is absolutely correct and solves the exercice, but it doesn't actually answer the convoluted question of the OP... (maybe this is a case for: math.meta.stackexchange.com/questions/30010/…)
  $endgroup$
  – Evargalo
  Apr 3 at 14:31
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Evargalo I agree that it is a good example for that discussion.
  $endgroup$
  – José Carlos Santos
  Apr 3 at 14:41
  

  
  
  
  

add a comment | 

4

$begingroup$

To remain on the safe side of strictness consider the partial sum

$$s(n) = sum_k=2^n left(frac1sqrtk-1-frac1sqrtk+1right)$$

and then look for the limit $ntoinfty$.

We have

$$s(n) = left( frac1sqrt1 +frac1sqrt2+frac1sqrt3+ ...+ frac1sqrtn-1right)\
-left( frac1sqrt3 +frac1sqrt4+ ...+ frac1sqrtn-1+frac1sqrtn+frac1sqrtn+1right)\
= 1+frac1sqrt2 -frac1sqrtn-frac1sqrtn+1 $$

and we get finally

$$lim_nto infty , s(n) =1+frac1sqrt2 $$

EDIT

The generalization to an arbitrary sequence $a(k)$ is easily done. For a fixed distance $d$ and for $n>d$ the partial sum is given by

$$s(d,n) = sum_k=1^n left( a_n-a_k+d right)= sum_j=1^d a_j-sum_j=1^d a_n+j$$

And if $lim_nto infty , a_n =0$ we find

$$s(d) = lim_nto infty , s(d,n) =sum_j=1^d a_j$$

share|cite|improve this answer

edited Apr 3 at 14:47

answered Apr 3 at 9:18

Dr. Wolfgang HintzeDr. Wolfgang Hintze

3,875621

$endgroup$

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
);



);

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%2f3172963%2fhow-to-write-the-general-formula-for-a-telescoping-series%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

Since$$sum_k=2^inftyleft(frac1sqrtk-1-frac1sqrtk+1right)=sum_k=2^inftyleft(frac1sqrtk-1-frac1sqrt kright)+sum_k=2^inftyleft(frac1sqrt k-frac1sqrtk+1right),$$the sum of your series is $1+dfrac1sqrt2$.

share|cite|improve this answer

answered Apr 3 at 8:57

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thgis is absolutely correct and solves the exercice, but it doesn't actually answer the convoluted question of the OP... (maybe this is a case for: math.meta.stackexchange.com/questions/30010/…)
  $endgroup$
  – Evargalo
  Apr 3 at 14:31
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Evargalo I agree that it is a good example for that discussion.
  $endgroup$
  – José Carlos Santos
  Apr 3 at 14:41
  

  
  
  
  

add a comment | 

2

$begingroup$

Since$$sum_k=2^inftyleft(frac1sqrtk-1-frac1sqrtk+1right)=sum_k=2^inftyleft(frac1sqrtk-1-frac1sqrt kright)+sum_k=2^inftyleft(frac1sqrt k-frac1sqrtk+1right),$$the sum of your series is $1+dfrac1sqrt2$.

share|cite|improve this answer

answered Apr 3 at 8:57

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thgis is absolutely correct and solves the exercice, but it doesn't actually answer the convoluted question of the OP... (maybe this is a case for: math.meta.stackexchange.com/questions/30010/…)
  $endgroup$
  – Evargalo
  Apr 3 at 14:31
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Evargalo I agree that it is a good example for that discussion.
  $endgroup$
  – José Carlos Santos
  Apr 3 at 14:41
  

  
  
  
  

add a comment | 

$begingroup$

Since$$sum_k=2^inftyleft(frac1sqrtk-1-frac1sqrtk+1right)=sum_k=2^inftyleft(frac1sqrtk-1-frac1sqrt kright)+sum_k=2^inftyleft(frac1sqrt k-frac1sqrtk+1right),$$the sum of your series is $1+dfrac1sqrt2$.

share|cite|improve this answer

answered Apr 3 at 8:57

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

share|cite|improve this answer

answered Apr 3 at 8:57

José Carlos SantosJosé Carlos Santos

173k23133241

answered Apr 3 at 8:57

José Carlos SantosJosé Carlos Santos

173k23133241

answered Apr 3 at 8:57

José Carlos SantosJosé Carlos Santos

173k23133241

4

$begingroup$

To remain on the safe side of strictness consider the partial sum

$$s(n) = sum_k=2^n left(frac1sqrtk-1-frac1sqrtk+1right)$$

and then look for the limit $ntoinfty$.

We have

$$s(n) = left( frac1sqrt1 +frac1sqrt2+frac1sqrt3+ ...+ frac1sqrtn-1right)\
-left( frac1sqrt3 +frac1sqrt4+ ...+ frac1sqrtn-1+frac1sqrtn+frac1sqrtn+1right)\
= 1+frac1sqrt2 -frac1sqrtn-frac1sqrtn+1 $$

and we get finally

$$lim_nto infty , s(n) =1+frac1sqrt2 $$

EDIT

The generalization to an arbitrary sequence $a(k)$ is easily done. For a fixed distance $d$ and for $n>d$ the partial sum is given by

$$s(d,n) = sum_k=1^n left( a_n-a_k+d right)= sum_j=1^d a_j-sum_j=1^d a_n+j$$

And if $lim_nto infty , a_n =0$ we find

$$s(d) = lim_nto infty , s(d,n) =sum_j=1^d a_j$$

share|cite|improve this answer

edited Apr 3 at 14:47

answered Apr 3 at 9:18

Dr. Wolfgang HintzeDr. Wolfgang Hintze

3,875621

$endgroup$

add a comment | 

4

$begingroup$

To remain on the safe side of strictness consider the partial sum

$$s(n) = sum_k=2^n left(frac1sqrtk-1-frac1sqrtk+1right)$$

and then look for the limit $ntoinfty$.

We have

$$s(n) = left( frac1sqrt1 +frac1sqrt2+frac1sqrt3+ ...+ frac1sqrtn-1right)\
-left( frac1sqrt3 +frac1sqrt4+ ...+ frac1sqrtn-1+frac1sqrtn+frac1sqrtn+1right)\
= 1+frac1sqrt2 -frac1sqrtn-frac1sqrtn+1 $$

and we get finally

$$lim_nto infty , s(n) =1+frac1sqrt2 $$

EDIT

The generalization to an arbitrary sequence $a(k)$ is easily done. For a fixed distance $d$ and for $n>d$ the partial sum is given by

$$s(d,n) = sum_k=1^n left( a_n-a_k+d right)= sum_j=1^d a_j-sum_j=1^d a_n+j$$

And if $lim_nto infty , a_n =0$ we find

$$s(d) = lim_nto infty , s(d,n) =sum_j=1^d a_j$$

share|cite|improve this answer

edited Apr 3 at 14:47

answered Apr 3 at 9:18

Dr. Wolfgang HintzeDr. Wolfgang Hintze

3,875621

$endgroup$

add a comment | 

$begingroup$

To remain on the safe side of strictness consider the partial sum

$$s(n) = sum_k=2^n left(frac1sqrtk-1-frac1sqrtk+1right)$$

and then look for the limit $ntoinfty$.

We have

$$s(n) = left( frac1sqrt1 +frac1sqrt2+frac1sqrt3+ ...+ frac1sqrtn-1right)\
-left( frac1sqrt3 +frac1sqrt4+ ...+ frac1sqrtn-1+frac1sqrtn+frac1sqrtn+1right)\
= 1+frac1sqrt2 -frac1sqrtn-frac1sqrtn+1 $$

and we get finally

$$lim_nto infty , s(n) =1+frac1sqrt2 $$

EDIT

The generalization to an arbitrary sequence $a(k)$ is easily done. For a fixed distance $d$ and for $n>d$ the partial sum is given by

$$s(d,n) = sum_k=1^n left( a_n-a_k+d right)= sum_j=1^d a_j-sum_j=1^d a_n+j$$

And if $lim_nto infty , a_n =0$ we find

$$s(d) = lim_nto infty , s(d,n) =sum_j=1^d a_j$$

share|cite|improve this answer

edited Apr 3 at 14:47

answered Apr 3 at 9:18

Dr. Wolfgang HintzeDr. Wolfgang Hintze

3,875621

$endgroup$

share|cite|improve this answer

edited Apr 3 at 14:47

answered Apr 3 at 9:18

Dr. Wolfgang HintzeDr. Wolfgang Hintze

3,875621

answered Apr 3 at 9:18

Dr. Wolfgang HintzeDr. Wolfgang Hintze

3,875621

answered Apr 3 at 9:18

Dr. Wolfgang HintzeDr. Wolfgang Hintze

3,875621