Proof of $\sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n}$ [duplicate]

Solution 1:

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 $n>1$.

Solution 2:

For inductive step :

$$\begin{align}\sum_{k=1}^{n+1}\frac{1}{k^2}&=\color{red}{\sum_{k=1}^{n}\frac{1}{k^2}}+\frac{1}{(n+1)^2}\\&\lt\color{red}{2-\frac 1n}+\frac{1}{(n+1)^2}\\&=2-\frac{(n+1)^2-n}{n(n+1)^2}\\&=2-\frac{1}{n+1}\left(\frac{n^2+n+1}{n(n+1)}\right)\\&=2-\frac{1}{n+1}\left(1+\frac{1}{n(n+1)}\right)\\&\lt 2-\frac{1}{n+1}\end{align}$$