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.

up vote 5 down vote accepted

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 at 20:00
  • 1
    @user376343 you'r re right, it was Leibniz. – Henno Brandsma Oct 14 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 $$ \int_{0}^{+\infty}\frac{(\sin x+\cos x)^2}{e^{2x/3}}\,dx,\qquad \int_{0}^{+\infty}\frac{e^{2x/3}}{(e^x+1)^2}\,dx.$$

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