I want to determine whether the following series converges:

$$\sum\limits_{k=1}^{\infty}\frac{(-1)^k k}{k^2-(-1)^k}$$

I managed to prove that it does not converge absolutely, however I fail to prove that the series converge conditionally (which seems to be the case). The Alternating series test doesn't work (because $|a_1|<|a_2|$).

  • 6
    $\begingroup$ While it is true that $|a_1|<|a_2|$, it is true that $|a_n|>|a_{n+1}|$ for $n$ large, so you can use the alternating sum test by excluding the first terms. $\endgroup$ – Thomas Andrews Jul 12 '16 at 19:13

Using alternating series test.

1) prove $\lim\limits_{n\to\infty}a_n=0$

$$a_n=\frac n{n^2-(-1)^n}=\frac {1/n}{1-\frac{(-1)^n}{n^2}} \rightarrow 0$$

2) prove $a_n \gt a_{n+1}$

$$a_n=\frac n{n^2-(-1)^n}=\frac 1{n-\frac{(-1)^n}{n}}$$

$$a_{n+1}=\frac 1{(n+1)-\frac{(-1)^{n+1}}{n+1}}=\frac 1{n-\frac{(-1)^n}n+ (1 -\frac{(-1)^{n+1}}{n+1}+\frac{(-1)^n}n)}$$

when $n \gt 10$

$$1 -\frac{(-1)^{n+1}}{n+1}+\frac{(-1)^n}n>1-1/10-1/10 =4/5$$

$$a_{n+1}=\frac 1{n-\frac{(-1)^n}n+ (1 -\frac{(-1)^{n+1}}{n+1}+\frac{(-1)^n}n)}<\frac 1{n-\frac{(-1)^n}n+ 4/5}<\frac 1{n-\frac{(-1)^n}n}=a_n$$


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$$\sum_{k=1}^{2N}\frac{(-1)^k k}{k^2-(-1)^k} = \sum_{h=1}^{N}\frac{2h}{4h^2-1}-\sum_{k=1}^{N}\frac{2h-1}{(2h-1)^2+1}\\=-\sum_{h=1}^{N}\frac{4h^2-6 h+1}{2 (2h-1) (2h+1) \left(2h^2-2 h+1\right)} $$ and the last one is a converging series, since the main term behaves like $\frac{1}{4h^2}$.
By partial fraction decomposition, the value of the series is given by: $$ -\frac{1}{2}+\sum_{n\geq 0}\frac{1}{2(2n+1)(2n^2+2n+1)}\approx 0.0477$$ that can be written as: $$ \left(\frac{7}{8}\zeta(3)-1\right)-\sum_{n\geq 1}\frac{1}{(2n+1)^3(4n^2+4n+2)}.$$

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