An infinite sum

$\dfrac{n}{n^4+n^2+1}$ can be factorized as $\dfrac{n}{(n^2+n+1)(n^2-n+1)}=\dfrac{1/2}{n^2-n+1}-\dfrac{1/2}{n^2+n+1}$

We assume that $a_n=\dfrac{1/2}{n^2-n+1}$

So, $a_{n+1}=\dfrac{1/2}{n^2+n+1}$

Hence,

$S_N=\displaystyle\sum_{n=0}^N \dfrac{n}{n^4+n^2+1}=\displaystyle\sum_{n=0}^N (a_n-a_{n+1})$

$=(a_0-a_1)+(a_1-a_2)+....+(a_N-a_{N+1})=a_0-a_{N+1}$

$=\dfrac{1}{2}-\dfrac{N}{N^2+N+1}$

Finally, $N \rightarrow \infty$

$S_{\infty}=\boxed {\dfrac{1}{2}}$

this is no proof, but:

the first few terms of the sequence are 1/3 , 2/21 , 3/91 , 4/273 ...

if we take the "cumulative sequence" of the sequence above, we get:

1/3 , (1/3 + 2/21) , (1/3 + 2/21 + 3/91), (1/3 + 2/21 + 3/91 + 4/273) ... = 1/3 , 3/7 , 6/13 , 10/21 ...

This sequence is the triangle numbers (formula n(n+1)/2) divided by the sequence n^2+n+1

The formula for this sequence is therefore n(n+1)/2(n^2+n+1) = (n^2+n)/(2n^2+2n+1)

The limit of this function (which is the answer) is 1/2, as this is the ratio between the coefficients of the dominant terms