Mathematical Induction Inequality problem I am trying to solve the following problem with  mathematical induction:
$$
\forall n>1,\qquad \frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<\frac{n-1}{n}
$$
but since I am new to the concept when it comes to inequalities I can't quite seem to work it out. 
Help, anyone?
 A: Let $$f(n) = \frac{1}{2^2} + \cdots + \frac{1}{n^2}.$$ Now we want to show that $$f(n)<\frac{n-1}{n}\tag{1}$$ for all integers greater than $1$. A proof by induction consists of two equally important steps. In the base case we show that $(1)$ indeed holds for $n=2$. In the inductive step we assume that $(1)$ is true for some number $n$ and use that to show that it is also true for $n+1$. 
Base case: We have $f(2)=\frac{1}{4} < \frac{1}{2}=\frac{2-1}{2}.$
Inductive step: Assume $(1)$ is true for some $n \geq 2$. We can calculate 
\begin{align*}
f(n+1) &= \color{green}{f(n)} + \frac{1}{(n+1)^2} \\
& < \color{green}{\frac{n-1}{n}} + \frac{1}{(n+1)^2} \\
& < \frac{n-1}{n} + \frac{1}{n\cdot(n+1)} \\
&= \frac{(n+1)-1}{n+1}.
\end{align*}
The expressions in green indicate the essential part of the inductive step. This shows that $(1)$ is true for $n+1$. Together with the base case this proves $(1)$ for all $n\geq 2$.
A: Hint $\ $ To prove $\,f_n = {\rm rhs} - {\rm lhs} > 0\,$ for all $\,n\ge 2\,$ note $\,f_2 > 0\,$ (base) $ $ and note 
$$\ \color{#c00}{f_{n+1}-f_n} =\, \frac{1}{n(n+1)}-\frac{1}{(n+1)^2}\, =\, \frac{1}{n(n+1)^2} \color{#c00}{> 0}$$
thus $\,  f_n > 0 \,\Rightarrow\, \color{#c00}{f_{n+1} > f_n} > 0\ $ (induction step)
Remark $\ $ The induction essentially shows that an increasing sequence stays $\ge $ its initial value, whose inductive proof is obvious, as above.  Note how rearranging the inequality into standard form $\, x > 0\,$ allowed us to simplify the induction into a more intuitive and more general form.  Many inductions can similarly be preprocessed to greatly simplify them. This is a special case of telescopic induction, about which you can find much discussion in prior posts on telescopy.
A: We want to prove:
$$ H_{n}^{(2)}\leq 2-\frac{1}{n}\tag{1} $$
by induction. $(1)$ holds for $n=1$, hence it is enough to prove that for every $n\geq 1$
$$ H_{n+1}^{(2)}-H_{n}^{(2)}=\frac{1}{(n+1)^2}\leq \frac{1}{n}-\frac{1}{n+1} = \frac{1}{n(n+1)}\tag{2} $$
holds, but that is trivial. In a similar way we me prove the improved inequality:
$$\boxed{\forall n\geq 1,\qquad H_{n}^{(2)}=\sum_{k=1}^{n}\frac{1}{k^2}\leq \zeta(2)-\frac{1}{\left(n+\frac{1}{2}\right)}+\frac{1}{12\left(n+\frac{1}{2}\right)^3}.} \tag{3}$$
It is interesting to point out that the last inequality can be used to prove Stirling's inequality.
