Evaluating $\sum\limits_{n=1}^\infty \frac{(-1)^{n+1}}{n}$

Question

I have this series: $$\sum_{n=1}^\infty s_n=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{(-1)^{n+1}}{n}+...$$

The sum of the first N terms is: $$S_N=\sum_{n=1}^N s_n=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{(-1)^{n+1}}{n}$$

It converges for Leibniz's test to a certain values $$\sigma=\sum_{n=1}^\infty s_n=\lim_{N\rightarrow \infty} S_N$$

I want to proof that $S_N \ge\frac{1}{2}$ and then $\sigma \ge\frac{1}{2}$ (in my book the author says that it is possible , but he doesn't show that)

$P(1)$ is true (because $s_1=1\ge \frac{1}{2}$).

We suppose that $P(N)$ is true (because $s_N\ge \frac{1}{2}$) and we have to proof that for $N+1$. $$S_{N+1}=\sum_{n=1}^N s_n+\frac{(-1)^{N+2}}{N+1}=S_N+\frac{(-1)^{N+2}}{N+1} \ge \frac{1}{2}+\frac{(-1)^{N+2}}{N+1}= \frac{N+1+2(-1)^{N+2}}{2*(N+1)} $$ If n is even $\frac{N+3}{2*(N+1)} \ge \frac {1}{2}$ If n is odd $\frac{N-1}{2*(N+1)} \le \frac {1}{2}$ where am I wrong?

Answer 1

HINT:

Note that

$$S_{2N}=\sum_{n=1}^{2N}\frac{(-1)^{n+1}}{n}=\sum_{n=1}^N \left(\frac{1}{2n-1}-\frac{1}{2n}\right)$$

and

$$S_{2N+1}=\sum_{n=1}^{2N+1}\frac{(-1)^{n+1}}{n}=\sum_{n=1}^N \left(\frac{1}{2n-1}-\frac{1}{2n}\right)+\frac1{2N+1}$$