An series using Basel Problem sum The Next CEO of Stack OverflowComputing a sum of an infinite series via partial decompositionComputing the sum of an infinite seriesInfinite sum of alternating telescoping seriesConvergence and sum of an infinite series: $sum_i=1^inftyfrac624 i-4 i^2-35$Fourier series, infinite seriesFind the power series of $f(x)=frac1x^2+x+1$Infinite series of formulaHow to prove the general formula for the Basel problem?Sum of the first n terms of series $fracx^3n3n(3n-1)(3n-2)$Find the sum of the infinite series $sum_n=3^infty [3/(n(n+3))]$
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An series using Basel Problem sum
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An series using Basel Problem sum
The Next CEO of Stack OverflowComputing a sum of an infinite series via partial decompositionComputing the sum of an infinite seriesInfinite sum of alternating telescoping seriesConvergence and sum of an infinite series: $sum_i=1^inftyfrac624 i-4 i^2-35$Fourier series, infinite seriesFind the power series of $f(x)=frac1x^2+x+1$Infinite series of formulaHow to prove the general formula for the Basel problem?Sum of the first n terms of series $fracx^3n3n(3n-1)(3n-2)$Find the sum of the infinite series $sum_n=3^infty [3/(n(n+3))]$
$begingroup$
I have this series I need to calculate:
$$sum_n=1^infty fracn+1n(n+2)^2$$
I've been trying to simplify the fraction via partial fraction decomposition, and I know I somehow need to make use of the Basel Problem, meaning that $$sum_n=1^infty frac1n^2 = fracpi^26$$ , but I don't know how to adress the movement of the index $n$ by $2$ in my original problem.
Any tips of how to proceed from here?
sequences-and-series summation
edited 6 mins ago
dmtri
1,7802521
asked 56 mins ago
Avi PAvi P
305
$endgroup$
add a comment |
$begingroup$
I have this series I need to calculate:
$$sum_n=1^infty fracn+1n(n+2)^2$$
I've been trying to simplify the fraction via partial fraction decomposition, and I know I somehow need to make use of the Basel Problem, meaning that $$sum_n=1^infty frac1n^2 = fracpi^26$$ , but I don't know how to adress the movement of the index $n$ by $2$ in my original problem.
Any tips of how to proceed from here?
sequences-and-series summation
edited 6 mins ago
dmtri
1,7802521
asked 56 mins ago
Avi PAvi P
305
$endgroup$
$begingroup$
Don't know if it helps, but what about $$sum_i=2^infty fraci(i-1)(i+1)^2=sum_i=2^infty fracii^3+i^2-i-1$$
$endgroup$
– Dr. Mathva
42 mins ago
2
$begingroup$
You seem to have forgotten a square in $$sum_n=1^infty frac1n^2 = fracpi^colorred26.$$
$endgroup$
– Servaes
32 mins ago
add a comment |
$begingroup$
I have this series I need to calculate:
$$sum_n=1^infty fracn+1n(n+2)^2$$
I've been trying to simplify the fraction via partial fraction decomposition, and I know I somehow need to make use of the Basel Problem, meaning that $$sum_n=1^infty frac1n^2 = fracpi^26$$ , but I don't know how to adress the movement of the index $n$ by $2$ in my original problem.
Any tips of how to proceed from here?
sequences-and-series summation
edited 6 mins ago
dmtri
1,7802521
asked 56 mins ago
Avi PAvi P
305
$endgroup$
I have this series I need to calculate:
$$sum_n=1^infty fracn+1n(n+2)^2$$
I've been trying to simplify the fraction via partial fraction decomposition, and I know I somehow need to make use of the Basel Problem, meaning that $$sum_n=1^infty frac1n^2 = fracpi^26$$ , but I don't know how to adress the movement of the index $n$ by $2$ in my original problem.
Any tips of how to proceed from here?
sequences-and-series summation
sequences-and-series summation
edited 6 mins ago
dmtri
1,7802521
asked 56 mins ago
Avi PAvi P
305
edited 6 mins ago
dmtri
1,7802521
asked 56 mins ago
Avi PAvi P
305
edited 6 mins ago
dmtri
1,7802521
edited 6 mins ago
dmtri
1,7802521
edited 6 mins ago
dmtri
1,7802521
1,7802521
asked 56 mins ago
Avi PAvi P
305
asked 56 mins ago
Avi PAvi P
305
asked 56 mins ago
Avi PAvi P
305
305
$begingroup$
Don't know if it helps, but what about $$sum_i=2^infty fraci(i-1)(i+1)^2=sum_i=2^infty fracii^3+i^2-i-1$$
$endgroup$
– Dr. Mathva
42 mins ago
2
$begingroup$
You seem to have forgotten a square in $$sum_n=1^infty frac1n^2 = fracpi^colorred26.$$
$endgroup$
– Servaes
32 mins ago
add a comment |
$begingroup$
Don't know if it helps, but what about $$sum_i=2^infty fraci(i-1)(i+1)^2=sum_i=2^infty fracii^3+i^2-i-1$$
$endgroup$
– Dr. Mathva
42 mins ago
2
$begingroup$
You seem to have forgotten a square in $$sum_n=1^infty frac1n^2 = fracpi^colorred26.$$
$endgroup$
– Servaes
32 mins ago
$begingroup$
Don't know if it helps, but what about $$sum_i=2^infty fraci(i-1)(i+1)^2=sum_i=2^infty fracii^3+i^2-i-1$$
$endgroup$
– Dr. Mathva
42 mins ago
$begingroup$
Don't know if it helps, but what about $$sum_i=2^infty fraci(i-1)(i+1)^2=sum_i=2^infty fracii^3+i^2-i-1$$
$endgroup$
– Dr. Mathva
42 mins ago
2
2
$begingroup$
You seem to have forgotten a square in $$sum_n=1^infty frac1n^2 = fracpi^colorred26.$$
$endgroup$
– Servaes
32 mins ago
$begingroup$
You seem to have forgotten a square in $$sum_n=1^infty frac1n^2 = fracpi^colorred26.$$
$endgroup$
– Servaes
32 mins ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Write (for example using Partial-fraction decomposition)
$$fracn+1n(n+2)^2 = frac12(n+2)^2+frac14left(frac1n-frac1n+2right).$$
The second bracket summed is a telescoping sum, specifically
$$sum_n=1^kfrac1n-frac1n+2=left(frac11-frac13right)+left(frac12-frac14right)+left(frac13-frac15right)+left(frac14-frac16right)+dots+frac1k-frac1k+2,$$
which after canceling terms gives
$$sum_n=1^kfrac1n-frac1n+2=1+frac12+frac1k-frac1k+2 to frac32,, textas kto infty.$$
For the first expression, notice
$$
sum_n=1^inftyfrac1(n+2)^2=sum_n=3^inftyfrac1n^2=sum_n=1^inftyfrac1n^2-1-frac14.
$$
Can you put these together and finish?
answered 36 mins ago
SilSil
5,64921745
$endgroup$
1
$begingroup$
yes, thank you very much. the last line was all I needed.
$endgroup$
– Avi P
34 mins ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Write (for example using Partial-fraction decomposition)
$$fracn+1n(n+2)^2 = frac12(n+2)^2+frac14left(frac1n-frac1n+2right).$$
The second bracket summed is a telescoping sum, specifically
$$sum_n=1^kfrac1n-frac1n+2=left(frac11-frac13right)+left(frac12-frac14right)+left(frac13-frac15right)+left(frac14-frac16right)+dots+frac1k-frac1k+2,$$
which after canceling terms gives
$$sum_n=1^kfrac1n-frac1n+2=1+frac12+frac1k-frac1k+2 to frac32,, textas kto infty.$$
For the first expression, notice
$$
sum_n=1^inftyfrac1(n+2)^2=sum_n=3^inftyfrac1n^2=sum_n=1^inftyfrac1n^2-1-frac14.
$$
Can you put these together and finish?
answered 36 mins ago
SilSil
5,64921745
$endgroup$
1
$begingroup$
yes, thank you very much. the last line was all I needed.
$endgroup$
– Avi P
34 mins ago
add a comment |
$begingroup$
Write (for example using Partial-fraction decomposition)
$$fracn+1n(n+2)^2 = frac12(n+2)^2+frac14left(frac1n-frac1n+2right).$$
The second bracket summed is a telescoping sum, specifically
$$sum_n=1^kfrac1n-frac1n+2=left(frac11-frac13right)+left(frac12-frac14right)+left(frac13-frac15right)+left(frac14-frac16right)+dots+frac1k-frac1k+2,$$
which after canceling terms gives
$$sum_n=1^kfrac1n-frac1n+2=1+frac12+frac1k-frac1k+2 to frac32,, textas kto infty.$$
For the first expression, notice
$$
sum_n=1^inftyfrac1(n+2)^2=sum_n=3^inftyfrac1n^2=sum_n=1^inftyfrac1n^2-1-frac14.
$$
Can you put these together and finish?
answered 36 mins ago
SilSil
5,64921745
$endgroup$
1
$begingroup$
yes, thank you very much. the last line was all I needed.
$endgroup$
– Avi P
34 mins ago
add a comment |
$begingroup$
Write (for example using Partial-fraction decomposition)
$$fracn+1n(n+2)^2 = frac12(n+2)^2+frac14left(frac1n-frac1n+2right).$$
The second bracket summed is a telescoping sum, specifically
$$sum_n=1^kfrac1n-frac1n+2=left(frac11-frac13right)+left(frac12-frac14right)+left(frac13-frac15right)+left(frac14-frac16right)+dots+frac1k-frac1k+2,$$
which after canceling terms gives
$$sum_n=1^kfrac1n-frac1n+2=1+frac12+frac1k-frac1k+2 to frac32,, textas kto infty.$$
For the first expression, notice
$$
sum_n=1^inftyfrac1(n+2)^2=sum_n=3^inftyfrac1n^2=sum_n=1^inftyfrac1n^2-1-frac14.
$$
Can you put these together and finish?
answered 36 mins ago
SilSil
5,64921745
$endgroup$
Write (for example using Partial-fraction decomposition)
$$fracn+1n(n+2)^2 = frac12(n+2)^2+frac14left(frac1n-frac1n+2right).$$
The second bracket summed is a telescoping sum, specifically
$$sum_n=1^kfrac1n-frac1n+2=left(frac11-frac13right)+left(frac12-frac14right)+left(frac13-frac15right)+left(frac14-frac16right)+dots+frac1k-frac1k+2,$$
which after canceling terms gives
$$sum_n=1^kfrac1n-frac1n+2=1+frac12+frac1k-frac1k+2 to frac32,, textas kto infty.$$
For the first expression, notice
$$
sum_n=1^inftyfrac1(n+2)^2=sum_n=3^inftyfrac1n^2=sum_n=1^inftyfrac1n^2-1-frac14.
$$
Can you put these together and finish?
answered 36 mins ago
SilSil
5,64921745
edited 34 mins ago
answered 36 mins ago
SilSil
5,64921745
answered 36 mins ago
SilSil
5,64921745
answered 36 mins ago
SilSil
5,64921745
5,64921745
1
$begingroup$
yes, thank you very much. the last line was all I needed.
$endgroup$
– Avi P
34 mins ago
add a comment |
1
$begingroup$
yes, thank you very much. the last line was all I needed.
$endgroup$
– Avi P
34 mins ago
1
1
$begingroup$
yes, thank you very much. the last line was all I needed.
$endgroup$
– Avi P
34 mins ago
$begingroup$
yes, thank you very much. the last line was all I needed.
$endgroup$
– Avi P
34 mins ago
add a comment |
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$begingroup$
Don't know if it helps, but what about $$sum_i=2^infty fraci(i-1)(i+1)^2=sum_i=2^infty fracii^3+i^2-i-1$$
$endgroup$
– Dr. Mathva
42 mins ago
2
$begingroup$
You seem to have forgotten a square in $$sum_n=1^infty frac1n^2 = fracpi^colorred26.$$
$endgroup$
– Servaes
32 mins ago