$\sum_{n=1}^{\infty}\frac{n} {n^2+1}z^n$
I calculate derivative $ S'(z) = \sum_{n=1}^{\infty}\frac{n^2} {n^2+1}z^{n-1}$ and later $ \sum_{n=1}^{\infty}\frac{n^2+1-1} {n^2+1}z^{n-1}$ so we have $ \sum_{n=1}^{\infty}\frac{n^2+1} {n^2+1}z^{n-1} - \sum_{n=1}^{\infty}\frac{1} {n^2+1}z^{n-1}$ and first series is easy, but how calculate second?
As Theo Bendit commented $$S=\sum_{n=1}^{\infty}\frac{n} {n^2+1}z^n$$ is not the must funny you could find.
Writing $$\frac{n} {n^2+1}=\frac{1}{2}\left(\frac{1}{ n+i}+\frac{1}{ n-i}\right)$$ you end with $$S=\frac{1+i}{4} z\, [\, _2F_1(1+i,2;2+i;z)-i \, _2F_1(1-i,2;2-i;z)]$$ where appears the Gaussian hypergeometric function.
If $z$ is real, $S$ is real only if $-\infty < z <1$