<s>[INST] Determining whether $\frac{1}{2^{2}} + \frac{2}{3{^2}} + ... +\frac{n}{(n+1)^{2}}$ is a Cauchy sequence. My attempt so far: If $n \leq m$, then
$a_{m} - a_{n} = \frac{n+1}{(n+2)^2}+\frac{n+2}{(n+3)^2}+...+\frac{m}{(m+1)^2} < \frac{n+1}{(n+2)(n+1)}+\frac{n+2}{(n+3)(n+2)}+...+\frac{m}{(m+1)m} = \frac{1}{(n+2)}+\frac{1}{(n+3)}+...+\frac{1}{(m+1)} < \frac{1}{n}+\frac{1}{n}+...+\frac{1}{m} \leq \frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}$.
A bit stuck here though. The hint in the book said it was not Cauchy.
Edit: If it's not Cauchy, then let me take a new approach..
$a_{m} - a_{n} = \frac{n+1}{(n+2)^2}+\frac{n+2}{(n+3)^2}+...+\frac{m}{(m+1)^2} \geq \frac{n+1}{(n+2)^2} + \frac{n+2}{(n+3)^2}+...+\frac{m}{(m+1)^2} \geq \frac{1}{(m+1)^2} + \frac{1}{(m+1)^2}+...+\frac{1}{(m+1)^2}$.
The $\frac{1}{(m+1)^2}$ occurs $m-n$ times here.
 [/INST] Note that we simply have $k\geq (k+1)/2$ for any integer, so
$$
\sum_{k=1}^n \frac{k}{(k+1)^2} \geq \sum_{k=1}^n \frac{\frac{1}{2}(k+1)}{(k+1)^2} = \sum_{k=1}^n \frac{1}{2} \frac{1}{k+1} = \frac{1}{2} \sum_{k=1}^n \frac{1}{k+1}
$$
and note that the right most expression is nothing but harmonic series starting from $k=2$, and according to basic analysis results, it diverges to $\infty$, therefore the series is not Cauchy $\mathbb{R}$ (Of course you can say it is if you are consider $\mathbb{C} \cup \{\infty\}$)
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