Calculate $sum_n=1^infty fracn+1n(n+2)^2$ using Basel Problem sumCalc the sum of $sum_k = 0^infty frac(-1)^kk sin(2k)$Convergence and sum of an infinite series: $sum_i=1^inftyfrac624 i-4 i^2-35$Does this sum converge: $sum_n=0^infty fracn!2^nprod_i=1^n(1+fraci2)$?Calculate $sumlimits_n=1^infty frac1(n+2)(n+4)^2$Elementary way to calculate the series $sumlimits_n=1^inftyfracH_nn2^n$Computing:$sum_n=0^inftyfrac3^nn!(n+3)$How to prove the general formula for the Basel problem?Sum of the first n terms of series $fracx^3n3n(3n-1)(3n-2)$Help summing the telescoping series $sum_n=2^inftyfrac1n^3-n$.Find the sum of the infinite series $sum_n=3^infty [3/(n(n+3))]$

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Calculate $sum_n=1^infty fracn+1n(n+2)^2$ using Basel Problem sum


Calc the sum of $sum_k = 0^infty frac(-1)^kk sin(2k)$Convergence and sum of an infinite series: $sum_i=1^inftyfrac624 i-4 i^2-35$Does this sum converge: $sum_n=0^infty fracn!2^nprod_i=1^n(1+fraci2)$?Calculate $sumlimits_n=1^infty frac1(n+2)(n+4)^2$Elementary way to calculate the series $sumlimits_n=1^inftyfracH_nn2^n$Computing:$sum_n=0^inftyfrac3^nn!(n+3)$How to prove the general formula for the Basel problem?Sum of the first n terms of series $fracx^3n3n(3n-1)(3n-2)$Help summing the telescoping series $sum_n=2^inftyfrac1n^3-n$.Find the sum of the infinite series $sum_n=3^infty [3/(n(n+3))]$













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







share|cite|improve this question











$endgroup$











  • $begingroup$
    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
    2 days ago















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







share|cite|improve this question











$endgroup$











  • $begingroup$
    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
    2 days ago













3












3








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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









YuiTo Cheng

2,1863937




2,1863937










asked 2 days ago









Avi PAvi P

325




325











  • $begingroup$
    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
    2 days 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
    2 days 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
2 days 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
2 days ago










1 Answer
1






active

oldest

votes


















5












$begingroup$

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
$$sum_n=1^kfrac1n-frac1n+2=1+frac12-frac1k+1-frac1k+2 to frac32,, textas kto infty.$$



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?






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    yes, thank you very much. the last line was all I needed.
    $endgroup$
    – Avi P
    2 days ago











Your Answer





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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

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
$$sum_n=1^kfrac1n-frac1n+2=1+frac12-frac1k+1-frac1k+2 to frac32,, textas kto infty.$$



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?






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    yes, thank you very much. the last line was all I needed.
    $endgroup$
    – Avi P
    2 days ago















5












$begingroup$

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
$$sum_n=1^kfrac1n-frac1n+2=1+frac12-frac1k+1-frac1k+2 to frac32,, textas kto infty.$$



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?






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    yes, thank you very much. the last line was all I needed.
    $endgroup$
    – Avi P
    2 days ago













5












5








5





$begingroup$

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
$$sum_n=1^kfrac1n-frac1n+2=1+frac12-frac1k+1-frac1k+2 to frac32,, textas kto infty.$$



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?






share|cite|improve this answer











$endgroup$



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
$$sum_n=1^kfrac1n-frac1n+2=1+frac12-frac1k+1-frac1k+2 to frac32,, textas kto infty.$$



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?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered 2 days ago









SilSil

5,67421745




5,67421745







  • 1




    $begingroup$
    yes, thank you very much. the last line was all I needed.
    $endgroup$
    – Avi P
    2 days ago












  • 1




    $begingroup$
    yes, thank you very much. the last line was all I needed.
    $endgroup$
    – Avi P
    2 days ago







1




1




$begingroup$
yes, thank you very much. the last line was all I needed.
$endgroup$
– Avi P
2 days ago




$begingroup$
yes, thank you very much. the last line was all I needed.
$endgroup$
– Avi P
2 days ago

















draft saved

draft discarded
















































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