Question

Series

Answer 1

\[\sum_{n=1}^{\infty}\frac{ 1 }{ \sum_{k=1}^{n}k^3 }\]

Answer 2

@ganeshie8 @Hero @waterineyes @paki @Lyrae @asnaseer @eliassaab @timo86m

Answer 3

\[\sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4}\]

Answer 4

very funny @waterineyes

Answer 5

What happened?

Answer 6

pressuming I knew that what's next ?

Answer 7

He he he.. :)

Answer 8

pressuming that I know that the expression can be written as os sum of partial fractions
(1/n^2)+(2/n)-(2/(n+1))+(1/(n+1)^2)
Still what's next?

Answer 9

Using the result that @waterineyes stated above, your series becomes:\[S=\sum_{n=1}^{\infty}\frac{ 1 }{ \sum_{k=1}^{n}k^3 }=\sum_{n=1}^{\infty}\frac{4}{n^2(n+1)^2}=4\sum_{n=1}^{\infty}\frac{1}{n^2}+\frac{2}{n+1}+\frac{1}{(n+1)^2}-\frac{2}{n}\]If we start listing the terms we get:\[S=4\left((\frac{1}{1}+\cancel{\frac{2}{2}}+\color{red}{\frac{1}{4}}-\frac{2}{1})+(\color{red}{\frac{1}{4}}+\cancel{\color{blue}{\frac{2}{3}}}+\color{green}{\frac{1}{9}}-\cancel{\frac{2}{2}})+...\right)\]Notice how some terms cancel (the \(\color{blue}{\frac{2}{3}}\) term will cancel with the corresponding term in the next expansion) and some combine (the \(\color{green}{\frac{1}{9}}\) will combine with the corresponding term in the next expansion). We can therefore write this as:\[S=4\left(\frac{1}{1}+2\sum_{n=1}^{\infty}\frac{1}{(n+1)^2}-\frac{2}{1}\right)=4\left(2\sum_{n=1}^{\infty}\frac{1}{(n+1)^2}-1\right)\]Hopefully you can complete this from here

Answer 10

@asnaseer I think you're answer is not suitable

Answer 11

uh ya its pretty much done