$\sum_{n=1}^{\infty}\frac{1}{n^2} <\frac{33}{20}$ using elementary inequalities

You only need a few more terms: $$\zeta(2)<1+\frac14+\frac19+\sum_{n=4}^\infty\frac1{n^2-1/4} =1+\frac14+\frac19+\frac27$$ which is already less than $33/20$.

Beuker-like integrals are really helpful. We have $\zeta(2)=2\int_{0}^{1}\frac{-\log x}{1+x}\,dx $ and $$ \int_{0}^{1}\frac{-\log(x)x^4(1-x)^2}{1+x}\,dx= -\frac{493}{150}+2\,\zeta(2). $$ The LHS is positive and bounded by $\frac{1}{120}$, hence $$ \zeta(2) \in\left(\frac{493}{300},\frac{659}{400}\right)$$ implies $\zeta(2)=\color{green}{1.64}\ldots$

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By creative telescoping™, the inequality $x\leq \text{arctanh}(x)$ for $x\in(0,1)$ and the Weierstrass product for the (hyperbolic)sine function we also have $$ \zeta(2) \leq \frac{205}{144}+\frac{1}{2}\log\left(\frac{36\sinh\pi}{85\pi}\right) $$ implying $\zeta(2)<1.645=\frac{329}{200}$.

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 $0 < c < 1$.