Proof of $\sum_{k=1}^n \left(\frac{1}{k^2}\right)\le \:\:2-\frac{1}{n}$ Proof of $\displaystyle\sum_{k=1}^n\left(\frac{1}{k^2}\right)\le \:\:2-\frac{1}{n}$
The following proof is from a book, however, there is something that I don't quite understand
for $k\geq  2$ we have:
(1): $\displaystyle\frac{1}{k^2}\le \frac{1}{k\left(k-1\right)}$
so
(2):$\displaystyle \sum _{k=1}^n\left(\frac{1}{k^2}\right)\le \sum _{k=2}^n\left(\frac{1}{k\left(k-1\right)}\right)$
(3):$\displaystyle \sum_{k=1}^n\left(\frac{1}{k^2}\right)\le \:1+\left(1-\frac{1}{n}\right)$
(4):$\displaystyle\sum_{k=1}^n\left(\frac{1}{k^2}\right)\le \:\:2-\frac{1}{n}$
I don't understand why in line (2), on the right, we start the sum at $k=2$ while
on the left, we start on $k=1$. Why start at $k=2$?
 A: You are right, that the inequality (2) is wrong. The correct inequality should be :
$\displaystyle \sum _{k=1}^n\left(\frac{1}{k^2}\right) = \displaystyle \sum _{k=2}^n\left(\frac{1}{k^2}\right) + 1 \le \sum _{k=2}^n\left(\frac{1}{k\left(k-1\right)}\right) + \mathbf{1}$
which is also consistent with (3), (4).
The summation has to start from $k=2$ on the right-hand side because otherwise you would get division by zero  $\frac{1}{1\cdot(1-1)} = \frac{1}{0}$
A: Indeed we have
$$\begin{aligned}\sum _{k=1}^n\left(\frac{1}{k^2}\right)&=1+\sum _{k=2}^n\left(\frac{1}{k^2}\right) \\
&\le 1+\sum _{k=2}^n\left(\frac{1}{k\left(k-1\right)}\right) \\
&=1+\sum _{k=2}^n\left(\frac1{k-1}-\frac1k\right) \\
&=2-\frac1n\end{aligned}$$
A: Note that $k(k-1)<k^2$ for $k>0$ but when $k=1$, $\dfrac{1}{n(n-1)}$ is not defined.
$\begin{aligned}\displaystyle \sum_{k=2}^{n} \frac{1}{k^2} &<\sum_{k=2}^{n} \frac{1}{k(k-1)} \\
 &= \sum_{k=2}^{n} \frac{1}{k-1}-\frac{1}{k} \\
&=1-\frac{1}{n} \\ &\Rightarrow \sum_{k=1}^{n} \frac{1}{k^2} 
=1+\sum_{k=2}^{n} \frac{1}{k^2} \\
&<1+1-\frac{1}{n} \\
&=2-\frac{1}{n} \end{aligned}$
