Does the series $\sum_{n=1}^\infty n^{(-1)^n-2}$ converge?

Does the series $\sum_{n=1}^\infty n^{(-1)^n-2}$ converge?

I tried this way:

$$\sum_{n=1}^\infty n^{(-1)^n-2} = \sum_{n=1}^\infty \frac 1n - \sum_{n=1}^\infty \frac1{n+2} + \sum_{n=1}^\infty \frac1{n^3} -\sum_{n=1}^\infty \frac 1{(n+1)^3}$$

The first one is harmonic series and therefore diverges, the second one diverges by comparison test with harmonic series, and the third and forth converge by comparison test with $\sum_{n=1}^\infty \frac 1{n^2}$.

May I conclude that the original series diverges as sum of convergent and divergent series ?

• It is not clear what your series is? Are you saying that the exponent in the denominator is $3$ for odd numbered terms and $1$ for even numbered terms? If so, then Sami Ben's answer is correct. But this doesn match with what you wrote when describing your try.... Commented Nov 14, 2013 at 9:13
• You cannot draw conclusions from having split up a series like that, unless everything is absolutely convergent. A conditionally convergent series can do very weird things when you start rearranging, and everything goes out the door with divergence. Commented Nov 14, 2013 at 9:15
• My series is $\sum_1^\infty n^{(-1)^n-2}$ I edited the question and title Commented Nov 14, 2013 at 9:21
• If I cannot draw conclusions from my try, what should I do to solve this question ? Commented Nov 14, 2013 at 11:14
• @user97484 See Sami's answer. Commented Nov 14, 2013 at 20:07

Hint: your series is greater than $$\sum_n \frac {1}{2n}$$ so conclude.
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ The difference of the harmonics $\large\tt converges$: \begin{align} &\color{#0000ff}{\large\sum_{n = 1}^{\infty}\pars{{1 \over n} - {1 \over n +2}}} = 2\sum_{n = 1}^{\infty}{1 \over n\pars{n + 2}} = 2\sum_{n = 0}^{\infty}{1 \over \pars{n + 1}\pars{n + 3}} = \Psi\pars{3} - \Psi\pars{1} \\[3mm]&= \Psi\pars{2} + {1 \over 2} - \Psi\pars{1} = \Psi\pars{1} + {1 \over 1} + {1 \over 2} - \Psi\pars{1} = \color{#0000ff}{\large {3 \over 2}} \end{align} $\Psi\pars{z}$ is the $\tt\mbox{digamma function}$.
• There's no need of that to prove above sum is 3/2. Note that $\sum\frac{1}{n}-\sum\frac{1}{n+2}=1+1/2$. All the other terms cancel out. Commented Nov 14, 2013 at 9:28
It is known that $$\sum_{k=1}^x\frac1k=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x}\right)$$and so the first part of your sum is just $$\frac32$$. The second part is clearly $$1$$ through a similar line of logic.