<s>[INST] How to prove that $a_{n} := \frac{n^2}{(n+1)^2}+\frac{(n+1)^2}{(n+2)^2}$ converges to $2$ when $n\to\infty$ by the definition? Just wondering if this is the correct way to approach and $\epsilon$ n proof for convergence.
I am starting with the sequence
$\frac{n^2}{(n+1)^2}+\frac{(n+1)^2}{(n+2)^2}$
First step,
\begin{align*}
    \left|\frac{n^2}{(n+1)^2}+\frac{(n+1)^2}{(n+2)^2} - 2 \right| 
    &= \left|\frac{(n+2)^{2}n^{2}+(n+1)^{4}-2(n+1)^{2}(n+2)^{2}}{(n+1)^{2}(n+2)^{2}} \right|< \epsilon\\
\end{align*}
since we have,
$(n+2)^{2}n^{2}+(n+1)^{4}-2(n+1)^{2}(n+2)^{2} < 2n^{4}(2n^{2}+4n+1)$
and,
$\frac{1}{(n+1)^{2}(n+2)^{2}}< \frac{1}{n^{4}}$
so,
$
    \left|\frac{(n+2)^{2}n^{2}+(n+1)^{4}-2(n+1)^{2}(n+2)^{2}}{(n+1)^{2}(n+2)^{2}} \right|
    < \left|\frac{2n^{4}(2n^{2}+4n+1)}{n^{4}}\right|
    <2(2n^{2}+4n+1)\\[1em]
$
This is where I'm getting stuck: Solving we get,
$\frac{-\sqrt{\epsilon+2}-2}{2}<n<\frac{\sqrt{\epsilon +2}-2}{2}$
For any $\epsilon >0$. Does this make sense? Should I only be taking the positive value for epsilon?
 [/INST] HINT
As a start, I would recommend you to split the argument into two parts. The first part corresponds to:
\begin{align*}
\frac{n^{2}}{(n + 1)^{2}} - 1 = -\frac{2n + 1}{(n + 1)^{2}} = \frac{1}{(n + 1)^{2}} - \frac{2}{n + 1}
\end{align*}
Similarly, the second part corresponds to:
\begin{align*}
\frac{(n + 1)^{2}}{(n + 2)^{2}} - 1 = -\frac{2n + 3}{(n + 2)^{2}} = \frac{1}{(n + 2)^{2}} - \frac{2}{n + 2}
\end{align*}
Letting $n\geq n_{\varepsilon}$, we arrive at the relation:
\begin{align*}
\left|\frac{n^{2}}{(n + 1)^{2}} + \frac{(n + 1)^{2}}{(n + 2)^{2}} - 2\right| & = \left|\frac{1}{(n + 1)^{2}} + \frac{1}{(n + 2)^{2}} - \frac{2}{n + 1} - \frac{2}{n + 2}\right|\\\\
& \leq \frac{1}{(n + 1)^{2}} + \frac{1}{(n + 2)^{2}} + \frac{2}{n + 1} + \frac{2}{n + 2}\\\\
& \leq \frac{1}{n + 1} + \frac{1}{n + 2} + \frac{2}{n + 1} + \frac{2}{n + 2}\\\\
& \leq \frac{1}{n} + \frac{1}{n} + \frac{2}{n} + \frac{2}{n}\\\\
& = \frac{6}{n} \leq \frac{6}{n_{\varepsilon}}
\end{align*}
Can you take it from here?
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