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?