What is the Solution to the Complex Sum \sum_{n>1} \frac{3n^2+1}{(n^3-n)^3}?

[anemone]

Evaluate \(\displaystyle \sum_{n>1} \frac{3n^2+1}{(n^3-n)^3}\).

 

[MarkFL]

Evaluate \(\displaystyle \sum_{n>1} \frac{3n^2+1}{(n^3-n)^3}\).

My solution:

Spoiler

Let's write the sum as:

\(\displaystyle S_n=\sum_{k=2}^{n}\left(\frac{3k^2+1}{\left(k^3-k\right)^3}\right)\)

Performing a partial fraction decomposition on the summand, and using the properties of sums, we obtain:

\(\displaystyle S_n=\frac{1}{2}\sum_{k=2}^{n}\left(\frac{1}{(k+1)^3}-\frac{2}{k^3}+\frac{1}{(k-1)^3}\right)+\frac{3}{2}\sum_{k=2}^{n}\left(\frac{1}{(k+1)^2}-\frac{1}{(k-1)^2}\right)+3\sum_{k=2}^{n}\left(\frac{1}{k+1}-\frac{2}{k}+\frac{1}{k-1}\right)\)

Let's now look at the 3 sums on the right in turn:

\(\displaystyle \sum_{k=2}^{n}\left(\frac{1}{(k+1)^3}-\frac{2}{k^3}+\frac{1}{(k-1)^3}\right)=1+\frac{1}{8}-\frac{1}{4}+\sum_{k=4}^{n-2}\left(\frac{1}{k^3}-\frac{2}{k^3}+\frac{1}{k^3}\right)+\frac{1}{(n+1)^3}+\frac{1}{n^3}-\frac{2}{n^3}=\frac{7}{8}+\frac{1}{(n+1)^3}-\frac{1}{n^3}\)

\(\displaystyle \sum_{k=2}^{n}\left(\frac{1}{(k+1)^2}-\frac{1}{(k-1)^2}\right)=-1-\frac{1}{4}+\sum_{k=4}^{n-2}\left(\frac{1}{k^2}-\frac{1}{k^2}\right)+\frac{1}{(n+1)^2}+\frac{1}{n^2}=-\frac{5}{4}+\frac{1}{(n+1)^2}+\frac{1}{n^2}\)

\(\displaystyle \sum_{k=2}^{n}\left(\frac{1}{k+1}-\frac{2}{k}+\frac{1}{k-1}\right)=1+\frac{1}{2}-\frac{2}{2}+\sum_{k=4}^{n-2}\left(\frac{1}{k}-\frac{2}{k}+\frac{1}{k}\right)+\frac{1}{n+1}+\frac{1}{n}-\frac{2}{n}=\frac{1}{2}+\frac{1}{n+1}-\frac{1}{n}\)

And so, we may now state the partial sum as:

\(\displaystyle S_n=\frac{1}{2}\left(\frac{7}{8}+\frac{1}{(n+1)^3}-\frac{1}{n^3}\right)+\frac{3}{2}\left(-\frac{5}{4}+\frac{1}{(n+1)^2}+\frac{1}{n^2}\right)+3\left(\frac{1}{2}+\frac{1}{n+1}-\frac{1}{n}\right)=\frac{(n(n+1))^3-8}{16(n(n+1))^3}\)

And so the infinite sum is:

\(\displaystyle S_{\infty}=\lim_{n\to\infty}\left(S_n\right)=\frac{1}{16}\)

 

[anemone]

My solution:

Spoiler

Let's write the sum as:

\(\displaystyle S_n=\sum_{k=2}^{n}\left(\frac{3k^2+1}{\left(k^3-k\right)^3}\right)\)

Performing a partial fraction decomposition on the summand, and using the properties of sums, we obtain:

\(\displaystyle S_n=\frac{1}{2}\sum_{k=2}^{n}\left(\frac{1}{(k+1)^3}-\frac{2}{k^3}+\frac{1}{(k-1)^3}\right)+\frac{3}{2}\sum_{k=2}^{n}\left(\frac{1}{(k+1)^2}-\frac{1}{(k-1)^2}\right)+3\sum_{k=2}^{n}\left(\frac{1}{k+1}-\frac{2}{k}+\frac{1}{k-1}\right)\)

Let's now look at the 3 sums on the right in turn:

\(\displaystyle \sum_{k=2}^{n}\left(\frac{1}{(k+1)^3}-\frac{2}{k^3}+\frac{1}{(k-1)^3}\right)=1+\frac{1}{8}-\frac{1}{4}+\sum_{k=4}^{n-2}\left(\frac{1}{k^3}-\frac{2}{k^3}+\frac{1}{k^3}\right)+\frac{1}{(n+1)^3}+\frac{1}{n^3}-\frac{2}{n^3}=\frac{7}{8}+\frac{1}{(n+1)^3}-\frac{1}{n^3}\)

\(\displaystyle \sum_{k=2}^{n}\left(\frac{1}{(k+1)^2}-\frac{1}{(k-1)^2}\right)=-1-\frac{1}{4}+\sum_{k=4}^{n-2}\left(\frac{1}{k^2}-\frac{1}{k^2}\right)+\frac{1}{(n+1)^2}+\frac{1}{n^2}=-\frac{5}{4}+\frac{1}{(n+1)^2}+\frac{1}{n^2}\)

\(\displaystyle \sum_{k=2}^{n}\left(\frac{1}{k+1}-\frac{2}{k}+\frac{1}{k-1}\right)=1+\frac{1}{2}-\frac{2}{2}+\sum_{k=4}^{n-2}\left(\frac{1}{k}-\frac{2}{k}+\frac{1}{k}\right)+\frac{1}{n+1}+\frac{1}{n}-\frac{2}{n}=\frac{1}{2}+\frac{1}{n+1}-\frac{1}{n}\)

And so, we may now state the partial sum as:

\(\displaystyle S_n=\frac{1}{2}\left(\frac{7}{8}+\frac{1}{(n+1)^3}-\frac{1}{n^3}\right)+\frac{3}{2}\left(-\frac{5}{4}+\frac{1}{(n+1)^2}+\frac{1}{n^2}\right)+3\left(\frac{1}{2}+\frac{1}{n+1}-\frac{1}{n}\right)=\frac{(n(n+1))^3-8}{16(n(n+1))^3}\)

And so the infinite sum is:

\(\displaystyle S_{\infty}=\lim_{n\to\infty}\left(S_n\right)=\frac{1}{16}\)

Awesome, MarkFL!(Cool)

 

[anemone]

Solution of other:

Spoiler

Note that $2(3n^2+1)=(n+1)^3-(n-1)^3$, so the sum becomes

\(\displaystyle \begin{align*}\sum_{n>1} \frac{3n^2+1}{(n^3-n)^3}&=\frac{1}{2}\sum_{n>1} \frac{(n+1)^3-(n-1)^3}{n^3(n+1)^3(n-1)^3}\\&=\frac{1}{2}\left(\sum_{n>1} \frac{1}{n^3(n-1)^3}-\sum_{n>1} \frac{1}{n^3(n+1)^3}\right)\\&=\frac{1}{2}\left(\frac{1}{2^3(2-1)^3}\right)\\&=\frac{1}{16}\end{align*}\)