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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!

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    $\begingroup$ May be you could separate the odd terms and the even terms. $\endgroup$ – Claude Leibovici Nov 27 '13 at 10:32
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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).

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  • $\begingroup$ 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] $\endgroup$ – Gaste Nov 27 '13 at 10:45
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    $\begingroup$ @Gaste: never trust a machine ;) Have you never read/seen science fiction? $\endgroup$ – Siméon Nov 27 '13 at 10:57
  • $\begingroup$ @Siméon. Very good point ! $\endgroup$ – Claude Leibovici Nov 27 '13 at 11:05
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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.

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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.

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ \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.

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