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!
 A: 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).
A: 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.
A: 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.
A: $\newcommand{\+}{^{\dagger}}%
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 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
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\begin{align}
&\sum_{n=0}^{\infty}\pars{-1}^{n}\,{2+ \pars{-1}^{n} \over n + 1}
=
\sum_{n=0}^{\infty}\bracks{{3 \over 2n + 1} - {1 \over 2n + 2}}
=
\sum_{n=0}^{\infty}{4n + 5 \over \pars{2n + 1}\pars{2n + 2}}
\end{align}
The 'serie general term' is $\quad\sim 1/n\quad$ when $\quad n \gg 1.\quad$The series $\color{#ff0000}{\large diverges}$ as the harmonic one.
