<s>[INST] Proof of $\sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n}$ Prove that for $n\geq 2, \: \sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n} $
I used induction and I compared the LHS and the RHS but I'm getting an incorrect inequality 
 [/INST] Meat of induction step:
\begin{align}
\sum_{i=1}^{k+1}\frac{1}{i^2} &= \frac{1}{(k+1)^2}+\sum_{i=1}^k\frac{1}{i^2}\\[1em]
&< \frac{1}{(k+1)^2}+2-\frac{1}{k}\tag{by ind. hyp}\\[1em]
&= 2-\frac{1}{k+1}\left(\frac{k+1}{k}-\frac{1}{k+1}\right)\\[1em]
&= 2-\frac{1}{k+1}\left(\frac{k^2-k}{k(k+1)}\right)\\[1em]
&< 2-\frac{1}{k+1}\tag{since $k\geq 2, k^2-k>0$}
\end{align}
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