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 $sum_r=0^∞ frac sin(r θ)3^r$
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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 $sum_r=0^∞ frac sin(r θ)3^r$
$begingroup$
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
edited May 14 at 10:25
YuiTo Cheng
3,25371245
asked May 14 at 8:06
Pranav AggarwalPranav Aggarwal
896
$endgroup$
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
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
edited May 14 at 10:25
YuiTo Cheng
3,25371245
asked May 14 at 8:06
Pranav AggarwalPranav Aggarwal
896
$endgroup$
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
summation
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
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
edited May 14 at 10:25
YuiTo Cheng
3,25371245
3,25371245
asked May 14 at 8:06
Pranav AggarwalPranav Aggarwal
896
asked May 14 at 8:06
Pranav AggarwalPranav Aggarwal
896
asked May 14 at 8:06
Pranav AggarwalPranav Aggarwal
896
896
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 |
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
2
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
$begingroup$
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
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
$begingroup$
Write it as $$1- sum_n=1^infty frac 18n-1-frac 18n+1$$
$endgroup$
– Mohammad Zuhair Khan
May 14 at 8:09
2
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
$begingroup$
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
$begingroup$
Related: math.stackexchange.com/questions/2553910/…
$endgroup$
– Greg Martin
May 14 at 8:30
4
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
$begingroup$
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 |
3 Answers
3
active
oldest
votes
$begingroup$
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.
answered May 14 at 8:41
Yves DaoustYves Daoust
137k877237
$endgroup$
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 $cotleft(fracpi 8right)=sqrt 2 +1$
answered May 14 at 9:20
Claude LeiboviciClaude Leibovici
128k1158137
$endgroup$
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
$$1-frac132sum_n=1^infty frac 1n^2-frac18^2=1-frac132frac1-fracpi8cot(pi/8)2(frac164)=frac1+sqrt28pi$$
answered May 14 at 22:18
E.H.EE.H.E
17.9k11969
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
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.
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.
answered May 14 at 8:41
Yves DaoustYves Daoust
137k877237
$endgroup$
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.
answered May 14 at 8:41
Yves DaoustYves Daoust
137k877237
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
137k877237
add a comment |
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 $cotleft(fracpi 8right)=sqrt 2 +1$
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 $cotleft(fracpi 8right)=sqrt 2 +1$
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 $cotleft(fracpi 8right)=sqrt 2 +1$
answered May 14 at 9:20
Claude LeiboviciClaude Leibovici
128k1158137
$endgroup$
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 $cotleft(fracpi 8right)=sqrt 2 +1$
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
answered May 14 at 9:20
Claude LeiboviciClaude Leibovici
128k1158137
128k1158137
add a comment |
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
$$1-frac132sum_n=1^infty frac 1n^2-frac18^2=1-frac132frac1-fracpi8cot(pi/8)2(frac164)=frac1+sqrt28pi$$
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
$$1-frac132sum_n=1^infty frac 1n^2-frac18^2=1-frac132frac1-fracpi8cot(pi/8)2(frac164)=frac1+sqrt28pi$$
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
$$1-frac132sum_n=1^infty frac 1n^2-frac18^2=1-frac132frac1-fracpi8cot(pi/8)2(frac164)=frac1+sqrt28pi$$
answered May 14 at 22:18
E.H.EE.H.E
17.9k11969
$endgroup$
$$1-dfrac 1 7+dfrac 1 9 - dfrac115 + dfrac 1 17+...=1- sum_n=1^infty frac 18n-1-frac 18n+1$$
$$=1-sum_n=1^infty frac 264n^2-1=1-frac132sum_n=1^infty frac 1n^2-frac18^2$$
and we have
$$frac1-pi x cot(pi x)2x^2=sum_n=1^infty frac1n^2-x^2$$
so
$$1-frac132sum_n=1^infty frac 1n^2-frac18^2=1-frac132frac1-fracpi8cot(pi/8)2(frac164)=frac1+sqrt28pi$$
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
answered May 14 at 22:18
E.H.EE.H.E
17.9k11969
17.9k11969
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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