Summation

We write $S=\displaystyle\sum_{n=1}^{99} t_n$ , where $t_n = \dfrac{n}{n^4 + n^2 + 1}$ . Then

$\begin{aligned} t_n &= \dfrac{n}{n^4 + n^2 + 1} \\ \\ &= \dfrac{n}{n^4 + 2n^2 + 1 - n^2} \\ \\ &= \dfrac{n}{(n^2 + 1)^2 - n^2} \\ \\ &= \dfrac{n}{(n^2 - n + 1)(n^2 + n + 1)} \\ \\ &= \dfrac{1}{2} \left( \dfrac{1}{n^2 - n + 1} - \dfrac{1}{n^2 + n + 1} \right) \end{aligned}$

Let $\ a_n = \dfrac{1}{n^2 - n + 1} \$ and $\ b_n = \dfrac{1}{n^2 + n + 1}$ , so that $\ t_n = \dfrac{1}{2} (a_n - b_n)$ . Consider $\ a_{n+1}$ .

$\begin{aligned} a_{n+1} &= \dfrac{1}{(n + 1)^2 - (n + 1) + 1} \\ \\ &= \dfrac{1}{n^2 + 2n + 1 - n - 1 + 1} \\ \\ &= \dfrac{1}{n^2 + n + 1} \\ \\ \\ &= b_n \end{aligned}$

which means that every $\text{-}b_k$ will cancel every $a_{k+1}$ . Thus we have

$\begin{aligned} S &= t_1 + t_2 + t_3 + .... + t_{98} + t_{99} \\ \\ &= \dfrac{1}{2} (a_1 - b_1) + \dfrac{1}{2} (a_2 - b_2) + \dfrac{1}{2} (a_3 - b_3) + .... + \dfrac{1}{2} (a_{98} - b_{98}) + \dfrac{1}{2} (a_{99} - b_{99}) \\ \\ &= \dfrac{1}{2} a_1 + \dfrac{1}{2} (\text{-}b_1 + a_2) + \dfrac{1}{2} (\text{-}b_2 + a_3) + \dfrac{1}{2} (\text{-}b_3 + a_4) + .... + \dfrac{1}{2} (\text{-}b_{97} + a_{98}) + \dfrac{1}{2} (\text{-}b_{98} + a_{99}) - \dfrac{1}{2} b_{99} \\ \\ &= \dfrac{1}{2} a_1 - \dfrac{1}{2} b_{99} \\ \\ &= \dfrac{1}{2} \left( \dfrac{1}{1^2 - 1 + 1} \right) - \dfrac{1}{2} \left( \dfrac{1}{99^2 + 99 + 1} \right) \\ \\ &= \dfrac{1}{2} - \dfrac{1}{19802} \end{aligned}$

which clearly falls between $0.49$ and $0.50$ .

Well my solution is similar @zico quintina however, not so clean and well explained . Here is the same problem telescoping series .