Find expression for : $ S_n =\sum_{i=1}^{n} \frac{i}{i^4+i^2+1} $ I want to find a formula for the sum of this series using its general term.
How to do it?
Series
$$
S_n = \underbrace{1/3 + 2/21 + 3/91 + 4/273 + \cdots}_{n \text{ terms}}
$$
General Term
$$
S_n = \sum_{i=1}^{n} \frac{i}{i^4+i^2+1}
$$
 A: Using partial fractions, we get $\dfrac{i}{i^4+i^2+1} = \dfrac{\tfrac{1}{2}}{i^2-i+1} - \dfrac{\tfrac{1}{2}}{i^2+i+1}$. 
Thus, $S_n = \displaystyle\sum_{i = 1}^{n}\dfrac{i}{i^4+i^2+1} = \dfrac{1}{2}\sum_{i = 1}^{n}\dfrac{1}{i^2-i+1} - \dfrac{1}{i^2+i+1}$. 
Since $(i+1)^2-(i+1)+1 = i^2+i+1$, this sum telescopes to $S_n = \dfrac{1}{2}\left(1 - \dfrac{1}{n^2+n+1}\right)$. 
Taking the limit as $n \to \infty$ gives $S = \dfrac{1}{2}$.
A: Use partial fractions,
$$ \frac{i}{i^4+i^2+1} = \frac{1}{2(i^2-i+1)}-\frac{1}{2(i^2+i+1)} $$
A: Write it out:
$$
\begin{eqnarray}
\sum_{i=1}^\infty \frac{i}{i^4 + i^2 + 1}
&=& \sum_{i=1}^\infty \left( \frac{2}{4i^2 - 4i + 4}
  - \frac{2}{4i^2 + 4i + 4}\right)\\
&=&  \sum_{i=1}^\infty \left( \frac{2}{\Big(2i-1\Big)^2 + 3}
  - \frac{2}{\Big( 2i + 1\Big)^2 + 3}\right)\\
&=&  \frac{2}{\Big( 2 - 1\Big)^2 + 3} + \sum_{i=1}^\infty \left( \frac{2}{\Big(2i+1\Big)^2 + 3}
  - \frac{2}{\Big( 2i + 1\Big)^2 + 3}\right)\\
&=& \frac{1}{2}.
\end{eqnarray}
$$
A: Set $$\color{blue}{a_k=  \frac{1}{k^2 -k +1} \implies a_{k+1}=  \frac{1}{(k+1)^2 -k } =\frac{1}{k^2+ k +1}} $$
But since $ (k^2 -k +1)(k^2 +k +1) =k^4 +k^2 +1$ we have,
$$ a_{k+1} -a_k = \frac{1}{k^2 +k +1}-  \frac{1}{k^2 -k +1} = \frac{2k}{(k^2 -k +1)(k^2 +k +1)} =\frac{2k}{k^4 +k^2 +1}$$
Then 
$$a_n-a_0 = \sum_{k=0}^{n-1}a_{k+1} -a_k=2\sum_{k=0}^{n-1}\frac{k}{k^4 +k^2 +1} $$
