algebra precalculus - Partial Fractions continued...

Hi asked the following question yesterday: Obtaining the sum of a series

Given the answers to that question by wj32, I am now trying to solve the following problem:

Consider the series
$$\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}$$
Use partial fractions to write the general term
$$u_n=\frac{n}{n^4+n^2+1}$$
as a difference of two simpler terms

My attempt at a Solution:
The partial fractions are
$$\begin{align}
\frac{n}{n^4+n^2+1}&=\frac{n+n-n}{n^4+n^2+1}\\
&=\frac{n+n}{n^4+n^2+1}-\frac{n}{n^4+n^2+1}\\
&=\frac{2}{n^3+n+\frac{1}{n}}-\frac{2}{2(n^3+n+\frac{1}{n})}
\end{align}$$
Then
$$\begin{align}
S_n&=\left[\sum_{n=1}^k\frac{2}{n^3+n+\frac{1}{n}}-\sum_{n=1}^k\frac{2}{2(n^3+n+\frac{1}{n})}\right]\\
&=\left[\left(\frac{2}{3}+\frac{4}{21}+\frac{6}{91}+...+\frac{2}{n^3+n+\frac{1}{n}}\right)-\left(\frac{1}{3}+\frac{2}{21}+\frac{3}{91}+...+\frac{2}{2(n^3+n+\frac{1}{n})}\right)\right]

\end{align}$$

Here the terms in the left do not cancel the terms in the right?

I'm guessing I need simpler/different partial fractions up at the top?

Answer

First note that we have a nice factorization of $n^4 + n^2 + 1$.
$$n^4 + n^2 + 1 =(n^2+n+1)(n^2-n+1)$$
Hence, $n = \dfrac{(n^2+n+1) - (n^2-n+1)}2$. This gives us $$u_n = \dfrac{(n^2+n+1) - (n^2-n+1)}{2(n^2+n+1)(n^2-n+1)} = \dfrac12 \left(\dfrac1{n^2-n+1} - \dfrac1{n^2+n+1}\right) = \dfrac12 \left(\dfrac1{(n-1)n+1} - \dfrac1{n(n+1)+1}\right)$$
Now telescopic summation should do the job for you

Hence,
\begin{align}
S_N = \sum_{n=1}^{N} u_n & = \dfrac12 \sum_{n=1}^N\left(\dfrac1{(n-1)n+1} - \dfrac1{n(n+1)+1}\right)
\end{align}
\begin{align}
2S_N & = \sum_{n=1}^N\left(\dfrac1{(n-1)n+1} - \dfrac1{n(n+1)+1}\right)\\
& = \left(1 - \dfrac13 \right) + \left(\dfrac13 - \dfrac17 \right) + \left(\dfrac17 - \dfrac1{13} \right) + \cdots\\
& + \left(\dfrac1{(n-2)(n-1)+1} - \dfrac1{(n-1)n+1}\right) + \left(\dfrac1{(n-1)n+1} - \dfrac1{n(n+1)+1}\right)\\
& = 1 - \dfrac1{n(n+1)+1}
\end{align}