convergency of the sequence $x_n=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+(-1)^{n+1}\frac{1}{n}.$ > Test the convergency of the **sequence** $\\{x_n\\}$ , where $$x_n=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+(-1)^{n+1}\frac{1}{n}.$$ I think the sequence $\\{x_n\\}$ is convergent. As, $$|x_{2n}-x_n|=\left|(-1)^{n+2}\frac{1}{n+1}+(-1)^{n+3}\frac{1}{n+2}+...+(-1)^{2n}\frac{1}{n+n}\right|$$ $$\le\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$$ $$\le\frac{1}{n+1}+\frac{1}{n+1}+...+\frac{1}{n+1}=\frac{n}{n+1}\to 1 \text{ as } n\to \infty$$ As, $|x_{2n}-x_n|\not <\epsilon$ taking $\epsilon =\frac{1}{2}$ , so the sequence is NOT Cauchy and hence NOT convergent. But I saw in a book that the sequence is a Cauchy sequence. I could not able to understand my fallacy.

You may just use the Dirichlet test: $$ \frac1n \geq \frac1{n+1}, $$ $$ \frac1n \to 0, \, \text{as}\,\, n \to +\infty, $$ $$ \left|\sum_1^n (-1)^{k-1}\right| \leq 1 $$ ensuring the **convergence** of the series $$ \sum_1^\infty \frac{(-1)^{n-1}}{n}. $$