<s>[INST] Prove $\sum_{n=1}^{\infty} \frac{2n+1}{n^2(n+1)^2}=1$ Use $f(n) = \frac{1}{n}$ and $f(n) = \frac{1}{n^2}$ when:

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*$\sum_{n=1}^{\infty}[f(n+1) - f(n)] = \lim_{n \to \infty} f(n) - f(1)$
to prove $\sum_{n=1}^{\infty} \frac{2n+1}{n^2(n+1)^2}=1$
What I have tried:
$$\sum_{n=1}^{\infty} \frac{2n+1}{n^2(n+1)^2} =\sum_{n=1}^{\infty}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2} \right)$$
Therefore;
$$\implies \lim_{n \to \infty}f(n) - f(1) = \lim_{n \to \infty}\frac{\left(\frac{1}{n^2}\right)}{n}-\frac{1}{(1+1)^2}$$
Though it does not look to me that this equals to 1, only hints on where do I go from here?
 [/INST] $f(n)-f(1)=\frac 1 {n^{2}}-1 \to -1$ so $\sum_{n=1}^{\infty}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2} \right)=-\sum_{n=1}^{\infty}\left(\frac{1}{(n+1)^2}-\frac{1}{n^2} \right) =-\lim [f(n)-1]=1$
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