Induction on summation inequality stuck on Induction step working on a fairly simple induction problem, but stuck when decomposing the m+1^st summation and bringing in the I.H to show it holds.
I am bad with formatting so bare with me.
$$\sum_{k=1}^n \frac1{k^2} \leq 2 - \frac1n$$
Base Case:
Set $n = 1$ and show that the inequality holds
IH:
Assume for all $n = m$, $\sum_{k=1}^m \frac1{k^2} \leq 2 - \frac1m$  holds
I.S:
$$\sum_{k=1}^{m+1} \frac1{k^2} \leq 2 - \frac1{m+1}$$
since we can decompose m+1 to m we get
$$\sum_{k=1}^m \frac1{k^2} + \frac1{m^2} \le 2 - \frac1m$$
then we can use I.H since we have $\sum_{k=1}^m \frac1{k^2} \le 2 - \frac1{m+1}$
so I am trying to show $2 - 1/(m+1) + 1/(m^2)$, but I am stuck here since I cannot manipulate the equation to come to the m+1st equation on the RHS

i think it must be
$$\sum_{k=1}^m\frac{1}{k^2}+\frac{1}{(m+1)^2}\le 2-\frac{1}{m+1}$$ and it must be
$$\le 2-\frac{1}{m}+ \frac{1}{(m+1)^2}\le 2-\frac{1}{m+1}$$
