Proof verification for proving $\forall n \ge 2, 1 + \frac1{2^2} + \frac1{3^2} + \cdots + \frac1{n^2} < 2 − \frac1n$ by induction 
Prove by mathematical induction: $\forall n \ge 2, 1 + \frac1{2^2} + \frac1{3^2} + \cdots + \frac1{n^2} < 2 − \frac1n$

Basis Step: (We want to show, $P(2)$, which is 1 + $\frac1{2^2}<2-\frac12$).
$\frac1{2^2}=\frac14$, and $2-\frac12=\frac32$
So, $\frac1{2^2}<2-\frac12$.
Inductive step: (We want to show, $\forall n\ge2,\mathrm P(n)\to\mathrm P(n+1)$.
Let $k\ge2$ be an integer, arbitrary & fixed.
(We want to show, $\mathrm P(k)\to\mathrm P(k+1)$).
(I.H.) Assume $1+\frac1{2^2}+\frac1{3^2}+\cdots+\frac1{k^2}<2-\frac1k$ ("$\mathrm P(k)$")
(We want to show $\mathrm P(k+1)$, which is $1+\frac1{2^2}+\frac1{3^2}+\cdots+\frac1{(k+1)^2}<2−\frac1{k+1}$)
$$\begin{align}
1+\frac1{2^2}+\frac1{3^2}+\cdots+\frac1{(k+1)^2}&<2−\frac1{k+1}+\frac1{(k+1)^2}\tag{by I.H.}\\
&=2-\frac1k+\frac1{k^2+2k+1}
\end{align}$$
(Thus, $\mathrm P(k)\to\mathrm P(k+1)$ for all $\mathrm k\ge2$
 Also, $\mathrm P(2)$, so by mathematical induction, $\forall k\ge2,\mathrm P(k)$)
Can someone look at my solution and see if it is correct? I'm not 100% confident with it.
 A: OK. It appears you are not interested in looking at other questions that pose the same question; rather, you solely want to know whether or not your own solution is correct. To this, I would say no. If I were your teacher, then I would probably give you 6/10 on this problem. Why?
Using the induction hypothesis, you should have
$$
\color{blue}{1+\frac{1}{4}+\cdots+\frac{1}{k^2}}+\frac{1}{(k+1)^2}< \color{blue}{2-\frac{1}{k}}+\frac{1}{(k+1)^2}\tag{1}
$$
and not
$$
1+\frac1{2^2}+\frac1{3^2}+\cdots+\frac1{(k+1)^2}<2−\frac1{k+1}+\frac1{(k+1)^2},\tag{2}
$$
as it currently stands. The induction hypothesis is used on the part highlighted in blue in $(1)$, but you did not use it properly in $(2)$. Finally, by your own admission, your goal is to end up with
$$
1+\frac1{2^2}+\frac1{3^2}+\cdots+\frac1{(k+1)^2}<2−\frac1{k+1},
$$
but all you did was write
$$
1+\frac1{2^2}+\frac1{3^2}+\cdots+\frac1{(k+1)^2}=2-\frac1k+\frac1{k^2+2k+1},
$$
and then you claimed you proved the result (something you obviously did not do). In $(1)$, your task is to reduce the right-hand side down into $2-\frac{1}{k+1}$, thus proving $P(k+1)$. You can do this like so:
\begin{align}
1+\frac{1}{4}+\cdots+\frac{1}{k^2}+\frac{1}{(k+1)^2}
&<2-\frac{1}{k}+\frac{1}{(k+1)^2}\quad\text{(by $P(k)$, the ind. hyp.)}\\[1em]
&= 2-\frac{1}{k+1}\left(\frac{k+1}{k}-\frac{1}{k+1}\right)\\[1em]
&= 2-\frac{1}{k+1}\left(\frac{k^2-k}{k(k+1)}\right)\\[1em]
&\leq 2-\frac{1}{k+1}.\qquad\text{(since $k\geq 1, k^2-k\geq 0$)}
\end{align}
Why could you never reach this end form? It's because you did not apply the induction hypothesis properly in $(2)$. 
A: If you are not looking for induction you can follow this path
$$0\le \frac{1}{k^2}\le \frac{1}{k(k-1)}$$
And after noting that $\frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}$ and the sum telescopes...
A: For the proof step, notice that $2-\frac{1}{k} <2 -\frac{1}{k+1}$, and after subtracting $\frac{1}{(k+1)^2}$, and then show that $2- \frac{1}{k}<2 - \frac{1}{k+1}(1+\frac{1}{k+1})$
