# 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.... – Jyrki Lahtonen Nov 14 '13 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. – zibadawa timmy Nov 14 '13 at 9:15
• My series is $\sum_1^\infty n^{(-1)^n-2}$ I edited the question and title – user97484 Nov 14 '13 at 9:21
• If I cannot draw conclusions from my try, what should I do to solve this question ? – user97484 Nov 14 '13 at 11:14
• @user97484 See Sami's answer. – zibadawa timmy Nov 14 '13 at 20:07

Hint: your series is greater than $$\sum_n \frac {1}{2n}$$ so conclude.
• 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. – Grobber Nov 14 '13 at 9:28