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

I have to show that the following series convergences:

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

I have tried the following:

• The alternating series test cannot be applied, since $\frac{2+(-1)^n}{n+1}$ is not monotonically decreasing.
• I tried splitting up the series in to series $\sum_{n=0}^{\infty}a_n = \sum_{n=0}^{\infty}(-1)^n \frac{2}{n+1}$ and $\sum_{n=0}^{\infty}b_n=\sum_{n=0}^{\infty}(-1)^n \frac{(-1)^n}{n+1}$. I proofed the convergence of the first series using the alternating series test, but then i realized that the second series is divergent.
• I also tried using the ratio test: for even $n$ the sequence converges to $\frac{1}{3}$, but for odd $n$ the sequence converges to $3$. Therefore the ratio is also not successful.

I ran out of ideas to show the convergence of the series.

Thanks in advance for any help!

• May be you could separate the odd terms and the even terms. – Claude Leibovici Nov 27 '13 at 10:32

It is not convergent. To see this, let $$a_n = (-1)^n\frac{2}{n+1},\qquad b_n =\frac{1}{n+1},\qquad c_n = a_n + b_n.$$ The series $\sum a_n$ is convergent by the alternating test.

We are interested in the convergence of $\sum c_n$. If $\sum c_n$ was convergent, then $\sum b_n = \sum c_n - \sum a_n$ would also be convergent, which is known to be false (divergence of the harmonic series).

• Thanks for your answer. I already tried this (see second point), but i was confused, since mathematica says this series is convergent: SumConvergence[(-1)^n*(2 + (-1)^n)/(n + 1), n] – Gaste Nov 27 '13 at 10:45
• @Gaste: never trust a machine ;) Have you never read/seen science fiction? – Siméon Nov 27 '13 at 10:57
• @Siméon. Very good point ! – Claude Leibovici Nov 27 '13 at 11:05

If you sum two successive terms (for indices $2n-1$ and $2n$), you get

$$\frac{3}{2n+1} - \frac{1}{2n} = \frac{6n-2n-1}{2n(2n+1)}= \frac{4n-1}{2n(2n+1)}$$

And its sum is not convergent, thus your series is not either.

You could look at the partial sums:

$$\sum_{n=1}^{N}(-1)^n \frac{2+(-1)^n}{n+1}=\frac{2}{1}-\frac{2}{2}+\frac{2}{3}-\frac{2}{4}+...+\frac{1}{1}+\frac{1}{2}+...=2\sum_{k \leq N+1 \text{ odd}} \frac{1}{k}$$ and this diverges.
