Prove that for all $n ≥ 1$, $\sum_{i=1}^{n}\frac{1}{i^2}≤2 - \frac{1}{n}$ 
Prove that for all $n ≥ 1 $ 
$$\sum_{i=1}^{n}\frac{1}{i^2}≤2 - \frac{1}{n}$$

My attempt: 
By induction.
Base case: $n = 1$
$$1 ≤ 2 - 1 = 1$$
Induction step:
Suppose it is true for $n$, i.e
$$\sum_{i=1}^{n}\frac{1}{i^2}≤2 - \frac{1}{n}$$
Adding $\frac{1}{n^2+n}$ to both sides gives
$$\sum_{i=1}^{n}\frac{1}{i^2} + \frac{1}{n^2+n} ≤2 - \frac{1}{n} + \frac{1}{n^2+n} \implies $$
$$\sum_{i=1}^{n}\frac{1}{i^2} + \frac{1}{n^2+n} ≤2 - \frac{1}{n+1}$$
Since $\frac{1}{n^2 + 2n + 1} < \frac{1}{n^2 + n}$, we have
$$\begin{align}
2 - \frac{1}{n+1} & ≥ \sum_{i=1}^{n}\frac{1}{i^2} + \frac{1}{n^2+n}\\
& ≥ \sum_{i=1}^{n}\frac{1}{i^2} + \frac{1}{n^2+2n+1} \\
& = \sum_{i=1}^{n}\frac{1}{i^2} + \frac{1}{(n+1)^2} \\
& = \sum_{i=1}^{n+1}\frac{1}{i^2}
\end{align}$$
$\Box$
Is it correct?

Proving that
$$\sum_{k=1}^{n} \frac{1}{k^2} <2-\frac{1}{n}.$$
Notice that $$\frac{1}{k^2} <\frac{1}{k(k-1)}=\frac{1}{k-1}- \frac{1}{k},~~~ k \ge 2.$$Starting the telescopic summing from $k=2$ we get
$$\sum_{k=2}^n \frac{1}{k^2} \le 1-\frac{1}{n} \implies \sum_{k=1}^{n} \frac{1}{k^2} <2-\frac{1}{n}.$$
