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Calculating the infinite sum $1-frac 1 7+frac 1 9 - frac115 + frac 1 17mp …=frac1+sqrt28pi$

Help proving $L(1, chi) = frac pi3 sqrt3 $Value of this infinite sumFinding $sum frac1n^2+7n+9$Finding an inverse trigonometric sumSummation of series with terms $U_n=frac1n^2-n+1 -frac1n^2+n+1$Find the sum of the infinite series $sum n(n+1)/n!$Find the domain of convergence of the series $sum^infty_n=1fracn!x^2nn^n(1+x^2n)$Sum of the first n terms of series $fracx^3n3n(3n-1)(3n-2)$An expression for the sum $sumlimits _k=1^n-1 k , (n-k)^2$Integrate $int_0^1fraclnx1+xdx$ using $sumfrac1k^2=fracpi ^26$Trigonometric Summation $sum_r=0^∞ frac sin(r θ)3^r$

1

1

$begingroup$

Prove that
$$1-dfrac 1 7+dfrac 1 9 - dfrac115 + dfrac 1 17mp ...=dfrac1+sqrt28pi$$

My attempt: I tried to break it into two series
$$(1+1/9+1/17+...)-(1/7+1/15+1/23+...)$$
But I don't know how to proceed. Any hints would be appreciated.

summation

share|cite|improve this question

edited May 14 at 10:25

YuiTo Cheng

3,25371245

asked May 14 at 8:06

Pranav AggarwalPranav Aggarwal

896

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Breaking into two series in that way can lead to $infty - infty$ as your original series is not absolutely convergent
  $endgroup$
  – Henry
  May 14 at 8:08
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Write it as $$1- sum_n=1^infty frac 18n-1-frac 18n+1$$
  $endgroup$
  – Mohammad Zuhair Khan
  May 14 at 8:09
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: change every "1" to an appropriate power of x and then differentiate or do some other operations.
  $endgroup$
  – Feng Shao
  May 14 at 8:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related: math.stackexchange.com/questions/2553910/…
  $endgroup$
  – Greg Martin
  May 14 at 8:30
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  To be a bit less cryptic, @FengShao is suggesting that you consider the series $x-x^7/7+x^9/9-x^15/15+cdots$ and relate it to a simpler series that you can work with by hand.
  $endgroup$
  – Greg Martin
  May 14 at 8:32
  

  
  
  
  

add a comment | 

1

1

$begingroup$

Prove that
$$1-dfrac 1 7+dfrac 1 9 - dfrac115 + dfrac 1 17mp ...=dfrac1+sqrt28pi$$

My attempt: I tried to break it into two series
$$(1+1/9+1/17+...)-(1/7+1/15+1/23+...)$$
But I don't know how to proceed. Any hints would be appreciated.

summation

share|cite|improve this question

edited May 14 at 10:25

YuiTo Cheng

3,25371245

asked May 14 at 8:06

Pranav AggarwalPranav Aggarwal

896

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Breaking into two series in that way can lead to $infty - infty$ as your original series is not absolutely convergent
  $endgroup$
  – Henry
  May 14 at 8:08
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Write it as $$1- sum_n=1^infty frac 18n-1-frac 18n+1$$
  $endgroup$
  – Mohammad Zuhair Khan
  May 14 at 8:09
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: change every "1" to an appropriate power of x and then differentiate or do some other operations.
  $endgroup$
  – Feng Shao
  May 14 at 8:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related: math.stackexchange.com/questions/2553910/…
  $endgroup$
  – Greg Martin
  May 14 at 8:30
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  To be a bit less cryptic, @FengShao is suggesting that you consider the series $x-x^7/7+x^9/9-x^15/15+cdots$ and relate it to a simpler series that you can work with by hand.
  $endgroup$
  – Greg Martin
  May 14 at 8:32
  

  
  
  
  

add a comment | 

$begingroup$

Prove that
$$1-dfrac 1 7+dfrac 1 9 - dfrac115 + dfrac 1 17mp ...=dfrac1+sqrt28pi$$

My attempt: I tried to break it into two series
$$(1+1/9+1/17+...)-(1/7+1/15+1/23+...)$$
But I don't know how to proceed. Any hints would be appreciated.

summation

share|cite|improve this question

edited May 14 at 10:25

YuiTo Cheng

3,25371245

asked May 14 at 8:06

Pranav AggarwalPranav Aggarwal

896

$endgroup$

share|cite|improve this question

edited May 14 at 10:25

YuiTo Cheng

3,25371245

asked May 14 at 8:06

Pranav AggarwalPranav Aggarwal

896

share|cite|improve this question

edited May 14 at 10:25

YuiTo Cheng

3,25371245

asked May 14 at 8:06

Pranav AggarwalPranav Aggarwal

896

edited May 14 at 10:25

YuiTo Cheng

3,25371245

edited May 14 at 10:25

YuiTo Cheng

3,25371245

asked May 14 at 8:06

Pranav AggarwalPranav Aggarwal

896

asked May 14 at 8:06

Pranav AggarwalPranav Aggarwal

896

3 Answers
3

active

oldest

votes

6

$begingroup$

Using the hints by Mohammad Zuhair Khan and Feng Shao, let

$$f(x):=1-sum_n=0^inftyleft(fracx^8n-18n-1-fracx^8n+18n+1right).$$

Then if we differentiate term-wise,

$$f'(x)=-sum_n=0^infty(x^8n-2-x^8n).$$

Using the geometric sum formula,

$$f'(x)=-fracx^61-x^8+fracx^81-x^8=-fracx^6(1-x^2)1-x^8.$$

Finally,

$$f(1)=1-int_0^1fracx^6(1-x^2)1-x^8dx.$$

https://www.wolframalpha.com/input/?i=integrate+x%5E6(1-x%5E2)%2F(1-x%5E8)+from+0+to+1

I see no easy way to solve the integral, other than by decomposition in simple fractions, which is tedious.

share|cite|improve this answer

edited May 14 at 10:02

answered May 14 at 8:41

Yves DaoustYves Daoust

137k877237

$endgroup$

add a comment | 

4

$begingroup$

Just for your curiosity.

Since you received good hints and a good answer, let me show how we could compute the partial sum
$$S_p=1- sum_n=1^p frac 18n-1-sum_n=1^pfrac 18n+1$$
It write
$$S_p=1+frac18 left(psi
left(p+frac98right)-psi left(p+frac78right)-psi left(frac98right)+psi left(frac78right)right)$$ where appears the digamma function.

Using the asymptotics and continuing with Taylor series for large values of $p$
$$S_p=frac pi8 cot left(fracpi 8right)+frac132 p-frac164
p^2+Oleft(frac1p^3right)$$

Computing
$$S_5=frac106748767459111928041225approx 0.953726$$ while the above truncated series would give
$$fracpi8 cot left(fracpi 8right)+frac91600approx 0.953684$$

Just remember that, using the half angle, $tan left(fracpi 8right)=sqrt 2 -1$ makes $cotleft(fracpi 8right)=sqrt 2 +1$

share|cite|improve this answer

answered May 14 at 9:20

Claude LeiboviciClaude Leibovici

128k1158137

$endgroup$

add a comment | 

0

$begingroup$

$$1-dfrac 1 7+dfrac 1 9 - dfrac115 + dfrac 1 17+...=1- sum_n=1^infty frac 18n-1-frac 18n+1$$
$$=1-sum_n=1^infty frac 264n^2-1=1-frac132sum_n=1^infty frac 1n^2-frac18^2$$
and we have
$$frac1-pi x cot(pi x)2x^2=sum_n=1^infty frac1n^2-x^2$$
so
$$1-frac132sum_n=1^infty frac 1n^2-frac18^2=1-frac132frac1-fracpi8cot(pi/8)2(frac164)=frac1+sqrt28pi$$

share|cite|improve this answer

answered May 14 at 22:18

E.H.EE.H.E

17.9k11969

$endgroup$

add a comment | 

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6

$begingroup$

Using the hints by Mohammad Zuhair Khan and Feng Shao, let

$$f(x):=1-sum_n=0^inftyleft(fracx^8n-18n-1-fracx^8n+18n+1right).$$

Then if we differentiate term-wise,

$$f'(x)=-sum_n=0^infty(x^8n-2-x^8n).$$

Using the geometric sum formula,

$$f'(x)=-fracx^61-x^8+fracx^81-x^8=-fracx^6(1-x^2)1-x^8.$$

Finally,

$$f(1)=1-int_0^1fracx^6(1-x^2)1-x^8dx.$$

https://www.wolframalpha.com/input/?i=integrate+x%5E6(1-x%5E2)%2F(1-x%5E8)+from+0+to+1

I see no easy way to solve the integral, other than by decomposition in simple fractions, which is tedious.

share|cite|improve this answer

edited May 14 at 10:02

answered May 14 at 8:41

Yves DaoustYves Daoust

137k877237

$endgroup$

add a comment | 

6

$begingroup$

Using the hints by Mohammad Zuhair Khan and Feng Shao, let

$$f(x):=1-sum_n=0^inftyleft(fracx^8n-18n-1-fracx^8n+18n+1right).$$

Then if we differentiate term-wise,

$$f'(x)=-sum_n=0^infty(x^8n-2-x^8n).$$

Using the geometric sum formula,

$$f'(x)=-fracx^61-x^8+fracx^81-x^8=-fracx^6(1-x^2)1-x^8.$$

Finally,

$$f(1)=1-int_0^1fracx^6(1-x^2)1-x^8dx.$$

https://www.wolframalpha.com/input/?i=integrate+x%5E6(1-x%5E2)%2F(1-x%5E8)+from+0+to+1

I see no easy way to solve the integral, other than by decomposition in simple fractions, which is tedious.

share|cite|improve this answer

edited May 14 at 10:02

answered May 14 at 8:41

Yves DaoustYves Daoust

137k877237

$endgroup$

add a comment | 

$begingroup$

Using the hints by Mohammad Zuhair Khan and Feng Shao, let

$$f(x):=1-sum_n=0^inftyleft(fracx^8n-18n-1-fracx^8n+18n+1right).$$

Then if we differentiate term-wise,

$$f'(x)=-sum_n=0^infty(x^8n-2-x^8n).$$

Using the geometric sum formula,

$$f'(x)=-fracx^61-x^8+fracx^81-x^8=-fracx^6(1-x^2)1-x^8.$$

Finally,

$$f(1)=1-int_0^1fracx^6(1-x^2)1-x^8dx.$$

https://www.wolframalpha.com/input/?i=integrate+x%5E6(1-x%5E2)%2F(1-x%5E8)+from+0+to+1

I see no easy way to solve the integral, other than by decomposition in simple fractions, which is tedious.

share|cite|improve this answer

edited May 14 at 10:02

answered May 14 at 8:41

Yves DaoustYves Daoust

137k877237

$endgroup$

share|cite|improve this answer

edited May 14 at 10:02

answered May 14 at 8:41

Yves DaoustYves Daoust

137k877237

answered May 14 at 8:41

Yves DaoustYves Daoust

137k877237

answered May 14 at 8:41

Yves DaoustYves Daoust

137k877237

4

$begingroup$

Just for your curiosity.

Since you received good hints and a good answer, let me show how we could compute the partial sum
$$S_p=1- sum_n=1^p frac 18n-1-sum_n=1^pfrac 18n+1$$
It write
$$S_p=1+frac18 left(psi
left(p+frac98right)-psi left(p+frac78right)-psi left(frac98right)+psi left(frac78right)right)$$ where appears the digamma function.

Using the asymptotics and continuing with Taylor series for large values of $p$
$$S_p=frac pi8 cot left(fracpi 8right)+frac132 p-frac164
p^2+Oleft(frac1p^3right)$$

Computing
$$S_5=frac106748767459111928041225approx 0.953726$$ while the above truncated series would give
$$fracpi8 cot left(fracpi 8right)+frac91600approx 0.953684$$

Just remember that, using the half angle, $tan left(fracpi 8right)=sqrt 2 -1$ makes $cotleft(fracpi 8right)=sqrt 2 +1$

share|cite|improve this answer

answered May 14 at 9:20

Claude LeiboviciClaude Leibovici

128k1158137

$endgroup$

add a comment | 

4

$begingroup$

Just for your curiosity.

Since you received good hints and a good answer, let me show how we could compute the partial sum
$$S_p=1- sum_n=1^p frac 18n-1-sum_n=1^pfrac 18n+1$$
It write
$$S_p=1+frac18 left(psi
left(p+frac98right)-psi left(p+frac78right)-psi left(frac98right)+psi left(frac78right)right)$$ where appears the digamma function.

Using the asymptotics and continuing with Taylor series for large values of $p$
$$S_p=frac pi8 cot left(fracpi 8right)+frac132 p-frac164
p^2+Oleft(frac1p^3right)$$

Computing
$$S_5=frac106748767459111928041225approx 0.953726$$ while the above truncated series would give
$$fracpi8 cot left(fracpi 8right)+frac91600approx 0.953684$$

Just remember that, using the half angle, $tan left(fracpi 8right)=sqrt 2 -1$ makes $cotleft(fracpi 8right)=sqrt 2 +1$

share|cite|improve this answer

answered May 14 at 9:20

Claude LeiboviciClaude Leibovici

128k1158137

$endgroup$

add a comment | 

$begingroup$

Just for your curiosity.

Since you received good hints and a good answer, let me show how we could compute the partial sum
$$S_p=1- sum_n=1^p frac 18n-1-sum_n=1^pfrac 18n+1$$
It write
$$S_p=1+frac18 left(psi
left(p+frac98right)-psi left(p+frac78right)-psi left(frac98right)+psi left(frac78right)right)$$ where appears the digamma function.

Using the asymptotics and continuing with Taylor series for large values of $p$
$$S_p=frac pi8 cot left(fracpi 8right)+frac132 p-frac164
p^2+Oleft(frac1p^3right)$$

Computing
$$S_5=frac106748767459111928041225approx 0.953726$$ while the above truncated series would give
$$fracpi8 cot left(fracpi 8right)+frac91600approx 0.953684$$

Just remember that, using the half angle, $tan left(fracpi 8right)=sqrt 2 -1$ makes $cotleft(fracpi 8right)=sqrt 2 +1$

share|cite|improve this answer

answered May 14 at 9:20

Claude LeiboviciClaude Leibovici

128k1158137

$endgroup$

share|cite|improve this answer

answered May 14 at 9:20

Claude LeiboviciClaude Leibovici

128k1158137

answered May 14 at 9:20

Claude LeiboviciClaude Leibovici

128k1158137

answered May 14 at 9:20

Claude LeiboviciClaude Leibovici

128k1158137

0

$begingroup$

$$1-dfrac 1 7+dfrac 1 9 - dfrac115 + dfrac 1 17+...=1- sum_n=1^infty frac 18n-1-frac 18n+1$$
$$=1-sum_n=1^infty frac 264n^2-1=1-frac132sum_n=1^infty frac 1n^2-frac18^2$$
and we have
$$frac1-pi x cot(pi x)2x^2=sum_n=1^infty frac1n^2-x^2$$
so
$$1-frac132sum_n=1^infty frac 1n^2-frac18^2=1-frac132frac1-fracpi8cot(pi/8)2(frac164)=frac1+sqrt28pi$$

share|cite|improve this answer

answered May 14 at 22:18

E.H.EE.H.E

17.9k11969

$endgroup$

add a comment | 

0

$begingroup$

$$1-dfrac 1 7+dfrac 1 9 - dfrac115 + dfrac 1 17+...=1- sum_n=1^infty frac 18n-1-frac 18n+1$$
$$=1-sum_n=1^infty frac 264n^2-1=1-frac132sum_n=1^infty frac 1n^2-frac18^2$$
and we have
$$frac1-pi x cot(pi x)2x^2=sum_n=1^infty frac1n^2-x^2$$
so
$$1-frac132sum_n=1^infty frac 1n^2-frac18^2=1-frac132frac1-fracpi8cot(pi/8)2(frac164)=frac1+sqrt28pi$$

share|cite|improve this answer

answered May 14 at 22:18

E.H.EE.H.E

17.9k11969

$endgroup$

add a comment | 

$begingroup$

$$1-dfrac 1 7+dfrac 1 9 - dfrac115 + dfrac 1 17+...=1- sum_n=1^infty frac 18n-1-frac 18n+1$$
$$=1-sum_n=1^infty frac 264n^2-1=1-frac132sum_n=1^infty frac 1n^2-frac18^2$$
and we have
$$frac1-pi x cot(pi x)2x^2=sum_n=1^infty frac1n^2-x^2$$
so
$$1-frac132sum_n=1^infty frac 1n^2-frac18^2=1-frac132frac1-fracpi8cot(pi/8)2(frac164)=frac1+sqrt28pi$$

share|cite|improve this answer

answered May 14 at 22:18

E.H.EE.H.E

17.9k11969

$endgroup$

share|cite|improve this answer

answered May 14 at 22:18

E.H.EE.H.E

17.9k11969

answered May 14 at 22:18

E.H.EE.H.E

17.9k11969

answered May 14 at 22:18

E.H.EE.H.E

17.9k11969