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 
 A: By induction assume it is true for $n$.
Then, by induction,
$\sum_{k = 1}^{n+1} \frac{1}{k^{2}} =\sum_{k = 1}^{n} \frac{1}{k^{2}} +\frac{1}{(n+1)^2} < 2-\frac{1}{n} +\frac{1}{(n+1)^2}$.
So to finish we need $ 2-\frac{1}{n} +\frac{1}{(n+1)^2} \leq 2 - \frac{1}{n+1}$,
or $\frac{1}{n+1} \leq \frac{1}{n}-\frac{1}{(n+1)^2}$, which is true for $n>1$. 
A: For inductive step : 
$$\begin{align}\sum_{k=1}^{n+1}\frac{1}{k^2}&=\color{red}{\sum_{k=1}^{n}\frac{1}{k^2}}+\frac{1}{(n+1)^2}\\&\lt\color{red}{2-\frac 1n}+\frac{1}{(n+1)^2}\\&=2-\frac{(n+1)^2-n}{n(n+1)^2}\\&=2-\frac{1}{n+1}\left(\frac{n^2+n+1}{n(n+1)}\right)\\&=2-\frac{1}{n+1}\left(1+\frac{1}{n(n+1)}\right)\\&\lt 2-\frac{1}{n+1}\end{align}$$
A: 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}
