TITLE: $\sum_{n=1}^{\infty}\frac{1}{n^2} <\frac{33}{20}$ using elementary inequalities
QUESTION [6 upvotes]: There are many ingenious ways for proving $$\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6} \approx 1.6449$$
Using the inequality $$\frac{1}{n^2} < \frac{1}{n^{2}-\frac{1}{4}} =\frac{1}{n-\frac{1}{2}}-\frac{1}{n+\frac{1}{2}}$$ we can see that $$\sum_{n=2}^{\infty}\frac{1}{n^2} <\frac{2}{3} \Rightarrow \zeta(2)<\frac{5}{3}=1.6666$$
Can we improve upon these bounds using elementary inequalities? Like is it possible to show (of course without assuming $\zeta(2)\approx 1.64449$) that $\zeta(2)<\frac{33}{20}$?
If there are much nicer bounds which follow using elementary inequalities I would be happy to see them.

REPLY [2 votes]: Here is a way
to get good
bounds for 
the tail of the sum.
I will show that
$\sum_{x=m}^{\infty} \dfrac1{x^2}
=\dfrac1{m}+\dfrac1{2(m-1)m}
-\dfrac1{3(m-1)m(m+1)}
-\dfrac{c}{(m-1)m(m+1)(m+2)}
$
where
$0 < c < 1$.
This method
can be extended
to higher order 
error terms,
but what I have done
is enough for me.
I am sure that
this was known
to Euler,
but this is
independent.
$\dfrac1{x^2}-\dfrac1{x(x+1)}
=\dfrac{1}{x^2(x+1)}
$
and
$\dfrac{1}{x^2(x+1)}-\dfrac{1}{(x-1)x(x+1)}
=\dfrac{-1}{(x-1)x^2(x+1)}
$
so
$\dfrac1{x^2}
=\dfrac1{x(x+1)}+\dfrac{1}{(x-1)x(x+1)}
-\dfrac1{(x-1)x^2(x+1)}
$.
Let
$p(x, n)
=\prod_{k=0}^{n-1} (x+k)
$.
$\begin{array}\\
\dfrac1{p(x, n)}-\dfrac1{p(x+1, n)}
&=\dfrac1{\prod_{k=0}^{n-1} (x+k)}-\dfrac1{\prod_{k=0}^{n-1} (x+1+k)}\\
&=\dfrac1{\prod_{k=0}^{n-1} (x+k)}-\dfrac1{\prod_{k=1}^{n} (x+k)}\\
&=\dfrac1{\prod_{k=1}^{n-1} (x+k)}\left(\dfrac1{x}-\dfrac1{x+n}\right)\\
&=\dfrac1{\prod_{k=1}^{n-1} (x+k)}
\dfrac{n}{x(x+n)}\\
&=\dfrac{n}{\prod_{k=0}^{n} (x+k)}\\
&=\dfrac{n}{p(x, n+1)}
\\
\end{array}\\
$
Therefore
$\sum_{x=m}^{\infty} \dfrac{n}{p(x, n+1)}
=\sum_{x=m}^{\infty} (\dfrac1{p(x, n)}-\dfrac1{p(x+1, n)})
=\dfrac1{p(m, n)}
$
or
$\sum_{x=m}^{\infty} \dfrac1{p(x, n+1)}
=\dfrac1{np(m, n)}
$.
We have
$\begin{array}\\
\dfrac1{x^2}
&=\dfrac1{x(x+1)}+\dfrac{1}{(x-1)x(x+1)}
-\dfrac1{(x-1)x^2(x+1)}\\
&=\dfrac1{p(x, 2)}+\dfrac1{p(x-1, 3)}
-\dfrac1{(x-1)x^2(x+1)}\\
\text{so}\\
\sum_{x=m}^{\infty} \dfrac1{x^2}
&=\sum_{x=m}^{\infty}\dfrac1{p(x, 2)}+\sum_{x=m}^{\infty}\dfrac1{p(x-1, 3)}
-\sum_{x=m}^{\infty}\dfrac1{(x-1)x^2(x+1)}\\
&=\dfrac1{p(m, 1)}+\dfrac1{2p(m-1, 2)}
-\sum_{x=m}^{\infty}\dfrac1{(x-1)x^2(x+1)}\\
&=\dfrac1{m}+\dfrac1{2(m-1)m}
-\sum_{x=m}^{\infty}\dfrac1{(x-1)x^2(x+1)}\\
\end{array}
$
Also
$\begin{array}\\
\dfrac1{p(x-1, 4)}
&=\dfrac1{(x-1)x(x+1)(x+2)}\\
&\lt \dfrac1{(x-1)x^2(x+1)}\\
&\lt \dfrac1{(x-2)(x-1)x(x+1)}\\
&=\dfrac1{p(x-2, 4)}\\
\end{array}
$
and
$\dfrac1{p(x-2, 4)}-\dfrac1{p(x-1, 4)}
=\dfrac{4}{p(x-1, 5)}
$
so that
$0
\lt \dfrac1{(x-2)(x-1)x(x+1)}-\dfrac1{p(x-2, 4)}
\lt \dfrac{4}{p(x-1, 5)}
$.
Therefore,
since
$\sum_{x=m}^{\infty} \dfrac1{p(x-2, 4)}
=\dfrac1{4p(m-2, 3)}
$
and
$\sum_{x=m}^{\infty} \dfrac1{p(x-1, 5)}
=\dfrac1{5p(m-1, 4)}
$
$\begin{array}\\
\sum_{x=m}^{\infty}\dfrac1{(x-1)x^2(x+1)}
&\gt \sum_{x=m}^{\infty} \dfrac1{p(x-1, 4)}\\
&= \dfrac1{3p(m-1, 3)}\\
&= \dfrac1{3(m-1)m(m+1)}\\
\end{array}
$
and
$\begin{array}\\
\sum_{x=m}^{\infty}\dfrac1{(x-1)x^2(x+1)}
&\lt \sum_{x=m}^{\infty} \dfrac1{p(x-2, 4)}\\
&\lt \sum_{x=m}^{\infty} \dfrac1{p(x-1, 4)}+\sum_{x=m}^{\infty} \dfrac{4}{p(x-1, 5)}\\
&= \dfrac1{3p(m-1, 3)}+\dfrac1{p(m-1, 4)}\\
&= \dfrac1{3(m-1)m(m+1)}+\dfrac1{(m-1)m(m+1)(m+2)}\\
\end{array}
$
so that
$\sum_{x=m}^{\infty} \dfrac1{x^2}
=\dfrac1{m}+\dfrac1{2(m-1)m}
-\dfrac1{3(m-1)m(m+1)}
= -\dfrac{c}{(m-1)m(m+1)(m+2)}
$
where
$0 < c < 1$.