Calculate $sum_{n=1}^infty frac{n+1}{n(n+2)^2}$ using Basel Problem sum












3












$begingroup$


I have this series I need to calculate:
$$sum_{n=1}^infty frac{n+1}{n(n+2)^2}$$
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 $$sum_{n=1}^infty frac{1}{n^2} = frac{pi^2}{6}$$ , 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







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  • $begingroup$
    Don't know if it helps, but what about $$sum_{i=2}^infty {frac{i}{(i-1)(i+1)^2}}=sum_{i=2}^infty {frac{i}{i^3+i^2-i-1}}$$
    $endgroup$
    – Dr. Mathva
    Mar 31 at 9:24
















3












$begingroup$


I have this series I need to calculate:
$$sum_{n=1}^infty frac{n+1}{n(n+2)^2}$$
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 $$sum_{n=1}^infty frac{1}{n^2} = frac{pi^2}{6}$$ , 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 $$sum_{i=2}^infty {frac{i}{(i-1)(i+1)^2}}=sum_{i=2}^infty {frac{i}{i^3+i^2-i-1}}$$
    $endgroup$
    – Dr. Mathva
    Mar 31 at 9:24














3












3








3





$begingroup$


I have this series I need to calculate:
$$sum_{n=1}^infty frac{n+1}{n(n+2)^2}$$
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 $$sum_{n=1}^infty frac{1}{n^2} = frac{pi^2}{6}$$ , 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:
$$sum_{n=1}^infty frac{n+1}{n(n+2)^2}$$
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 $$sum_{n=1}^infty frac{1}{n^2} = frac{pi^2}{6}$$ , 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






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share|cite|improve this question













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share|cite|improve this question








edited Mar 31 at 11:12









YuiTo Cheng

2,40641037




2,40641037










asked Mar 31 at 9:10









Avi PAvi P

396




396












  • $begingroup$
    Don't know if it helps, but what about $$sum_{i=2}^infty {frac{i}{(i-1)(i+1)^2}}=sum_{i=2}^infty {frac{i}{i^3+i^2-i-1}}$$
    $endgroup$
    – Dr. Mathva
    Mar 31 at 9:24


















  • $begingroup$
    Don't know if it helps, but what about $$sum_{i=2}^infty {frac{i}{(i-1)(i+1)^2}}=sum_{i=2}^infty {frac{i}{i^3+i^2-i-1}}$$
    $endgroup$
    – Dr. Mathva
    Mar 31 at 9:24
















$begingroup$
Don't know if it helps, but what about $$sum_{i=2}^infty {frac{i}{(i-1)(i+1)^2}}=sum_{i=2}^infty {frac{i}{i^3+i^2-i-1}}$$
$endgroup$
– Dr. Mathva
Mar 31 at 9:24




$begingroup$
Don't know if it helps, but what about $$sum_{i=2}^infty {frac{i}{(i-1)(i+1)^2}}=sum_{i=2}^infty {frac{i}{i^3+i^2-i-1}}$$
$endgroup$
– Dr. Mathva
Mar 31 at 9:24










1 Answer
1






active

oldest

votes


















5












$begingroup$

Write (for example using Partial-fraction decomposition)
$$frac{n+1}{n(n+2)^2} = frac{1}{2(n+2)^2}+frac{1}{4}left(frac{1}{n}-frac{1}{n+2}right).$$
The second bracket summed is a telescoping sum, specifically
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=left(frac{1}{1}-frac{1}{3}right)+left(frac{1}{2}-frac{1}{4}right)+left(frac{1}{3}-frac{1}{5}right)+left(frac{1}{4}-frac{1}{6}right)+dots+frac{1}{k}-frac{1}{k+2},$$
which after canceling terms gives
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=1+frac{1}{2}-frac{1}{k+1}-frac{1}{k+2} to frac{3}{2},, text{as }kto infty.$$



For the first expression, notice
$$
sum_{n=1}^inftyfrac{1}{(n+2)^2}=sum_{n=3}^inftyfrac{1}{n^2}=sum_{n=1}^inftyfrac{1}{n^2}-1-frac{1}{4}.
$$
Can you put these together and finish?






share|cite|improve this answer











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  • 1




    $begingroup$
    yes, thank you very much. the last line was all I needed.
    $endgroup$
    – Avi P
    Mar 31 at 9:32












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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)
$$frac{n+1}{n(n+2)^2} = frac{1}{2(n+2)^2}+frac{1}{4}left(frac{1}{n}-frac{1}{n+2}right).$$
The second bracket summed is a telescoping sum, specifically
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=left(frac{1}{1}-frac{1}{3}right)+left(frac{1}{2}-frac{1}{4}right)+left(frac{1}{3}-frac{1}{5}right)+left(frac{1}{4}-frac{1}{6}right)+dots+frac{1}{k}-frac{1}{k+2},$$
which after canceling terms gives
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=1+frac{1}{2}-frac{1}{k+1}-frac{1}{k+2} to frac{3}{2},, text{as }kto infty.$$



For the first expression, notice
$$
sum_{n=1}^inftyfrac{1}{(n+2)^2}=sum_{n=3}^inftyfrac{1}{n^2}=sum_{n=1}^inftyfrac{1}{n^2}-1-frac{1}{4}.
$$
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
    Mar 31 at 9:32
















5












$begingroup$

Write (for example using Partial-fraction decomposition)
$$frac{n+1}{n(n+2)^2} = frac{1}{2(n+2)^2}+frac{1}{4}left(frac{1}{n}-frac{1}{n+2}right).$$
The second bracket summed is a telescoping sum, specifically
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=left(frac{1}{1}-frac{1}{3}right)+left(frac{1}{2}-frac{1}{4}right)+left(frac{1}{3}-frac{1}{5}right)+left(frac{1}{4}-frac{1}{6}right)+dots+frac{1}{k}-frac{1}{k+2},$$
which after canceling terms gives
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=1+frac{1}{2}-frac{1}{k+1}-frac{1}{k+2} to frac{3}{2},, text{as }kto infty.$$



For the first expression, notice
$$
sum_{n=1}^inftyfrac{1}{(n+2)^2}=sum_{n=3}^inftyfrac{1}{n^2}=sum_{n=1}^inftyfrac{1}{n^2}-1-frac{1}{4}.
$$
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
    Mar 31 at 9:32














5












5








5





$begingroup$

Write (for example using Partial-fraction decomposition)
$$frac{n+1}{n(n+2)^2} = frac{1}{2(n+2)^2}+frac{1}{4}left(frac{1}{n}-frac{1}{n+2}right).$$
The second bracket summed is a telescoping sum, specifically
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=left(frac{1}{1}-frac{1}{3}right)+left(frac{1}{2}-frac{1}{4}right)+left(frac{1}{3}-frac{1}{5}right)+left(frac{1}{4}-frac{1}{6}right)+dots+frac{1}{k}-frac{1}{k+2},$$
which after canceling terms gives
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=1+frac{1}{2}-frac{1}{k+1}-frac{1}{k+2} to frac{3}{2},, text{as }kto infty.$$



For the first expression, notice
$$
sum_{n=1}^inftyfrac{1}{(n+2)^2}=sum_{n=3}^inftyfrac{1}{n^2}=sum_{n=1}^inftyfrac{1}{n^2}-1-frac{1}{4}.
$$
Can you put these together and finish?






share|cite|improve this answer











$endgroup$



Write (for example using Partial-fraction decomposition)
$$frac{n+1}{n(n+2)^2} = frac{1}{2(n+2)^2}+frac{1}{4}left(frac{1}{n}-frac{1}{n+2}right).$$
The second bracket summed is a telescoping sum, specifically
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=left(frac{1}{1}-frac{1}{3}right)+left(frac{1}{2}-frac{1}{4}right)+left(frac{1}{3}-frac{1}{5}right)+left(frac{1}{4}-frac{1}{6}right)+dots+frac{1}{k}-frac{1}{k+2},$$
which after canceling terms gives
$$sum_{n=1}^kfrac{1}{n}-frac{1}{n+2}=1+frac{1}{2}-frac{1}{k+1}-frac{1}{k+2} to frac{3}{2},, text{as }kto infty.$$



For the first expression, notice
$$
sum_{n=1}^inftyfrac{1}{(n+2)^2}=sum_{n=3}^inftyfrac{1}{n^2}=sum_{n=1}^inftyfrac{1}{n^2}-1-frac{1}{4}.
$$
Can you put these together and finish?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 31 at 18:54

























answered Mar 31 at 9:30









SilSil

5,65721745




5,65721745








  • 1




    $begingroup$
    yes, thank you very much. the last line was all I needed.
    $endgroup$
    – Avi P
    Mar 31 at 9:32














  • 1




    $begingroup$
    yes, thank you very much. the last line was all I needed.
    $endgroup$
    – Avi P
    Mar 31 at 9:32








1




1




$begingroup$
yes, thank you very much. the last line was all I needed.
$endgroup$
– Avi P
Mar 31 at 9:32




$begingroup$
yes, thank you very much. the last line was all I needed.
$endgroup$
– Avi P
Mar 31 at 9:32


















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