Proving inequality $\ 1+\frac14+\frac19+\cdots+\frac1{n^2}\le 2-\frac1n$ using induction Question:

Prove $$\ 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}\le 2-\frac{1}{n},
\text{ for all natural } n$$

My attempt:
Base Case: $n=1$ is true:
I.H: Suppose $1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{k^2}\le 2-\frac{1}{k},$ for some natural $k.$
Now we prove true for $n = k+1$
$$ 1+\frac{1}{4}+\cdots+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}\le 2-\frac{1}{k}+\frac{1}{\left(k+1\right)^2},\text{ by induction hypothesis} $$
Now how do I show that $2-\frac{1}{k}+\frac{1}{\left(k+1\right)^2}\le 2-\frac{1}{\left(k+1\right)}\text{ ?}$
Have I done everything correctly up until here? 
If yes, how do I show this inequality is true? 
Any help would be appreciated. 
 A: You are right!
We need to prove that:
$$\frac{1}{(k+1)^2}<\frac{1}{k}-\frac{1}{k+1}$$ or
$$\frac{1}{(k+1)^2}<\frac{1}{k(k+1)},$$
which is obvious.
A: You may use creative telescoping for proving a much tighter inequality.
If we set $H_n^{(2)}=\sum_{k=1}^{n}\frac{1}{k^2}$, for any $n\geq 3$ we have:
$$\begin{eqnarray*} H_n^{(2)} &=& \sum_{k=1}^{n}\frac{1}{k(k+1)}+\sum_{k=1}^{n}\frac{1}{k^2(k+1)}\\&=&\left(1-\frac{1}{n+1}\right)+\frac{1}{2}+\sum_{k=1}^{n-1}\frac{1}{(k+1)^2 (k+2)}\\&=&\frac{3}{2}-\frac{1}{n+1}+\sum_{k=1}^{n-1}\frac{1}{k(k+1)(k+2)}-\sum_{k=1}^{n-1}\frac{1}{k(k+1)^2(k+2)}\\&=&\frac{5}{3}-\frac{2n+1}{2n(n+1)}-\sum_{k=1}^{n-2}\frac{1}{(k+1)(k+2)^2(k+3)}\\ &\color{red}{\leq}&\color{red}{\frac{5}{3}-\frac{1}{n}} \end{eqnarray*}$$
and the given inequality for $n\in\{1,2\}$ can be easily checked by hand.
It is interesting to point out that this approach leads to a short proof of Stirling's inequality.

As an alternative, the convexity of $\frac{1}{x^2}$ over $\mathbb{R}^+$ leads, by the Hermite-Hadamard inequality, to:
$$ H_{n}^{(2)}=\frac{\pi^2}{6}-\sum_{m\geq n+1}\frac{1}{m^2}\color{red}{\leq} \frac{\pi^2}{6}-\int_{n+1}^{+\infty}\frac{dx}{x^2}=\frac{\pi^2}{6}-\frac{1}{n+1} $$
$$ H_{n}^{(2)}\color{blue}{\geq} \frac{\pi^2}{6}-\frac{1}{n+\frac{1}{2}}. $$
