Algebra , Polynomials

Note that $\begin{aligned}\dfrac{n}{n^4+n^2+1}&= \dfrac{n}{n^4+2n^2+1-n^2}\\ &= \dfrac{n}{(n^2+1)^2-n^2}\\ &= \dfrac{n}{(n^2-n+1)(n^2+n+1)}\\ &= \dfrac{\dfrac{1}{n^2-n+1}-\dfrac{1}{n^2+n+1}}{2}\end{aligned}$

Now note that $\dfrac{1}{(n+1)^2-(n+1)+1}=\dfrac{1}{n^2+n+1}$

Thus, the series $\sum_{n=0}^{\infty}\dfrac{\dfrac{1}{n^2-n+1}-\dfrac{1}{n^2+n+1}}{2}$ is a telescoping one.

Thus, $\sum_{n=0}^{\infty}\dfrac{\dfrac{1}{n^2-n+1}-\dfrac{1}{n^2+n+1}}{2}=\dfrac{\dfrac{1}{1^2-1+1}}{2}=\boxed{\dfrac{1}{2}}$ and we are done.

I started by looking at the first few partial sums: $\frac{1}{3} , \frac{3}{7} , \frac{6}{13} , \frac{10}{21} , ...$

It's pretty clear that the partial sums are approaching $\frac{1}{2}$ but we need to prove this.

For each partial sum $S_k$ , if we look at the values of $\frac{1}{2} - S_k$ , we get $\frac{1}{6}, \frac{1}{14}, \frac{1}{26}, \frac{1}{42}, ...$

The numerators seem to always be $1$ , and the denominators are $6,14,26,42,...$ . The second differences in these numbers are constant (namely 4), which means they fit a quadratic equation (with leading coefficient 2). Suppose the denominators are of the form $2x^2 + ax + b$ . Plugging in $x = 1$ gives $2 + a + b = 6$ , and plugging in $x = 2$ gives $8 + 2a + b = 14$ . Therefore $a = 2$ and $b = 2$ .

Therefore, the pattern appears to be $\frac{1}{2} - S_k = \frac{1}{2k^2 + 2k + 2}$ , or $S_k = \frac{k^2 + k}{2(k^2 + k + 1)}$ .

We see that this formula holds true for the first few cases, so naturally we use induction to prove it is always true.

Suppose $S_k = \frac{k^2 + k}{2(k^2 + k + 1)}$ for some $k \in \mathbb{N}$ . Then, $S_{k+1} = S_k + \frac{k+1}{(k+1)^4 + (k+1)^2 + 1}$ $= \frac{k^2 + k}{2(k^2 + k + 1)} + \frac{k+1}{((k+1)^2 + 1)^2 - (k+1)^2}$ $= \frac{k^2 + k}{2(k^2 + k + 1)} + \frac{k+1}{((k+1)^2 + (k+1) + 1)((k+1)^2 - (k+1) + 1)}$ $= \frac{k^2 + k}{2(k^2 + k + 1)} + \frac{k+1}{(k^2+3k+3)(k^2+k+1)}$ $= \frac{(k^2 + k)(k^2 + 3k + 3) + 2(k+1)}{2(k^2+3k+3)(k^2+k+1)}$ $= \frac{(k+1)(k^3+3k^2+3k+2)}{2(k^2+3k+3)(k^2+k+1)}$ $= \frac{(k+1)(k+2)(k^2+k+1)}{2(k^2+3k+3)(k^2+k+1)}$ $= \frac{(k+1)^2+(k+1)}{2((k+1)^2 + (k+1) + 1)}$

Therefore, by induction, we have that $S_k = \frac{k^2 + k}{2(k^2 + k + 1)}$ for all $k \in \mathbb{N}$ . Therefore,

$\lim_{k \to \infty} S_k = \lim_{k \to \infty} \frac{k^2 + k}{2(k^2 + k + 1)} = \lim_{k \to \infty} \left(\frac{1}{2} - \frac{1}{2k^2 + 2k + 2} \right) = \frac{1}{2}$

Hence the answer is $\boxed{\frac{1}{2}}$ .