Proving using Mathematical Induction I'm Stuck on the last step, this is proving using mathematical induction, a lecture from my Elementary number theory class.
The question goes to,
Prove that $\sum_{k=1}^n \frac{1}{k^2}=\frac{1}{1^2}+\frac{1}{2^2}+\cdots+\frac{1}{n^2}\le 2-\frac{1}{n}$ whenever $n$ is a positive integer.
This is my Attempt,
Step 1: Base Case ($n=1$)
$$\sum_{k=1}^n \frac{1}{k^2}=\frac{1}{1^2}\le2-\frac{1}{1}$$
$$=1\le1$$, Therefore Base case is true.
Step 2: Induction Hypothesis
Suppose $\sum_{k=1}^n \frac{1}{k^2}=\frac{1}{n^2}\le2-\frac{1}{n}$ is true for $n=m$
$\implies$ $\sum_{k=1}^m \frac{1}{k^2}=\frac{1}{m^2}\le2-\frac{1}{m}$ $\forall m \in \mathbb{N}$
Step 3: $n = m+1$
$\sum_{k=1}^{m+1} \frac{1}{k^2}=\sum_{k=1}^m \frac{1}{k^2}+\frac{1}{(m+1)^2}\le2-\frac{1}{m}+\frac{1}{(m+1)^2}$
I'm Stuck on this step
 A: There seem to be quite a bit of typoes in this proof. Steps (2) and (3) aren't correct. First, it's false for any $n > 1$ that
$$\sum_{k=1}^{n}\frac{1}{k^{2}} = \frac{1}{n^{2}}.$$
Second, your upper summation bound in Step (3) can't depend on the index $k$, so I'm assuming you meant to type $n+1$ instead. Even supposing you meant to type it that way, it's false that
$$\sum_{k=1}^{n+1}\frac{1}{\left(k+1\right)^{2}} = \frac{1}{k^{2}}.$$
It's also confusing when you say $n=k$ in Step (2) and then say $n=k+1$ in Step (3).
(Answer) To point in the right direction, our inductive step should start with something along the lines of,
"Suppose for a fixed $n=m$ that
$$\sum_{k=1}^{m}\frac{1}{k^{2}}\le2-\frac{1}{m}."$$
Then you use the sum $\sum_{k=1}^{m+1}\frac{1}{k^{2}}$ and manipulate it such that the next steps will involve our inductive hypothesis.
A: Starting from the induction hypothesis which is $\sum_{k=1}^m \frac{1}{k^2}\le2-\frac{1}{m}$:- 
Thus:-
$$\sum_{k=1}^m \frac{1}{k^2}\le2-\frac{1}{m}$$
$$\sum_{k=1}^{m+1} \frac{1}{k^2}\le2-\frac{1}{m}+\frac{1}{(m+1)^{2}}$$
As $m\ge1$ so:-
$$\Rightarrow m^2+2m\le m^2+2m+1$$
$$\Rightarrow \frac{m(m+2)}{(m+1)^2}\le 1$$
$$\Rightarrow \frac{(m+1)+1}{(m+1)^2}\le \frac{1}{m}$$
$$\Rightarrow \frac{1}{(m+1)^2} + \frac{1}{m+1}\le \frac{1}{m}$$
$$\Rightarrow \frac{1}{(m+1)^2} -\frac{1}{m}  \le -\frac{1}{m+1}$$
$$\Rightarrow 2-\frac{1}{m}+\frac{1}{(m+1)^2} \le 2 -\frac{1}{m+1}$$
Now making use of the 2nd equation:-
$$\sum_{k=1}^{m+1} \frac{1}{k^2}\le2-\frac{1}{m}+\frac{1}{(m+1)^{2}}\le2 -\frac{1}{m+1}$$
Thus completing the induction step.
