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.

• Determine if what ? – Yves Daoust Oct 13 '18 at 12:27

It converges by Leibniz' criterion. $$|a_n| \rightarrow 0$$ decreasingly, and alternating signs. Absolutely, compare it to the harmonic $$\frac1n$$ series.

• Leibniz or Euler? – user376343 Oct 13 '18 at 20:00
• @user376343 you'r re right, it was Leibniz. – Henno Brandsma Oct 14 '18 at 5:17

With positive signs, the terms are of order $$n^{-1}$$ and the series diverges.

With alternating signs, the pairs of terms are of order $$n^{-2}$$ and the series converges.

A trick with often works. We note that the absolute value of the term is asymptotic to $$1/n$$. Then write: $$(-1)^n\frac{n+1}{n^2+1} = \frac{(-1)^n}{n} + b_n$$ and note that $$\sum\frac{(-1)^n}{n}$$ converges, and $$\sum b_n$$ converges absolutely.

Here, $$|b_n| = \frac{n-1}{(n^2+1)n}$$ so $$\sum|b_n|$$ converges by comparison with $$\sum\frac{1}{n^2}$$.

We consider consecutive n (even), n+1 (odd) term,

$$\displaystyle \frac{n+1}{n^2+1} - \frac{n+2}{(n+1)^2+1} = \frac{n(n+3)}{(n^2+1)(n^2+2n+1} \sim \frac{1}{n^2}$$

$$\sum_{n=1}^{n=\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$ which is convergent.

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 $$\int_{0}^{+\infty}\frac{(\sin x+\cos x)^2}{e^{2x/3}}\,dx,\qquad \int_{0}^{+\infty}\frac{e^{2x/3}}{(e^x+1)^2}\,dx.$$