Determine if $\sum_{n=0}^{\infty}(-1)^n\frac{n+1}{n^2+1}$ converges or diverges I'm having trouble figuring this one out.

$$\sum_{n=0}^{\infty} (-1)^n\frac{n+1}{n^2+1}$$

I think this is conditionally converging as it has $(-1)^n$ so we should take $\lvert(-1)^n\rvert$? I'm a little lost on this one.
Any help would be appreciated.

It is trivially convergent by Leibniz' test, an not absolutely convergent by asymptotic comparison with the harmonic series. Convergent to what? is a more interesting question. 
We may notice that
$$ \frac{n+1}{n^2+1} = \int_{0}^{+\infty} e^{-nx}\left(\sin x+\cos x\right)\,dx $$
hence
$$ \sum_{n\geq 0}\frac{n+1}{n^2+1}(-1)^n = 1-\int_{0}^{+\infty}\frac{\sin x+\cos x}{e^x+1}\,dx \approx 0.366404$$
which can be written in terms of $1,\frac{\pi}{\sinh \pi}$ and the digamma function $\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}$ evaluated at $\pm\frac{i}{2}$ and $\frac{1\pm i}{2}$. By the Cauchy-Schwarz inequality we have that $\int_{0}^{+\infty} \left(\sin x+\cos x\right)\frac{dx}{e^x+1}\,dx$ is not too far from $\sqrt{\frac{7}{10}}$, since
$$ \int_{0}^{+\infty}\frac{(\sin x+\cos x)^2}{e^x}\,dx=\frac{7}{5},\qquad \int_{0}^{+\infty}\frac{e^x}{(e^x+1)^2}\,dx=\frac{1}{2}.$$
A better bound can be derived by considering
$$ \int_{0}^{+\infty}\frac{(\sin x+\cos x)^2}{e^{2x/3}}\,dx,\qquad  \int_{0}^{+\infty}\frac{e^{2x/3}}{(e^x+1)^2}\,dx.$$
