# Proof verification of $x_n = {1\over2^2}+{2\over3^2}+\cdots+{n\over(n+1)^2}$ diverges by the negation of Cauchy

Let $$\{x_n\}$$ denote a sequence: $$x_n = {1\over2^2}+{2\over3^2}+\cdots+{n\over(n+1)^2}$$ Show that $$\{x_n\}$$ diverges using the negation of Cauchy criterion.

I would like to kindly request a verification of the below.

Start with the negation of the definition of Cauchy criterion: $$\exists \epsilon > 0\ \forall N\in\Bbb N: \exists n, m > N \land |x_n-x_m| \ge \epsilon$$

First let $$m>n$$, then: \begin{align} |x_n - x_m| = |x_m-x_n| &= \left|\sum_{k=n+1}^m\frac{k}{(k+1)^2}\right| \stackrel{x_k>0}{=} \\ &= \sum_{k=n+1}^m\frac{k}{(k+1)^2} \ge \\ &\ge \sum_{k=n+1}^m \frac{k}{(2k)^2} = \\ &=\sum_{k=n+1}^m \frac{1}{4k} \end{align}

Chose $$m = 2n$$: \begin{align} \sum_{k=n+1}^{m} \frac{1}{4k} &= \sum_{k=n+1}^{2n} \frac{1}{4k} \ge \\ &\ge \sum_{k=n+1}^{2n} \frac{1}{8n} = \\ &= n\cdot\frac{1}{8n} = \\ &= {1\over 8} \end{align} Thus: $$|x_n - x_m| \ge {1\over 8}$$

Chose $$\epsilon = {1\over 8}$$: $$\exists \epsilon = {1\over 8}\ \forall N \in \Bbb N: \exists n, m>N \land |x_n - x_m| \ge {1\over 8}$$

Have i missed something?

• It looks right to me. – saulspatz Dec 20 '18 at 15:22