Upper bound for $\displaystyle\int_1^\infty \frac{\lfloor x\rfloor }{x^3} dx $ How do I find $f(x) > \lfloor x \rfloor $ in the interval $[1, \infty)$ such that 
$$\int_1^\infty \frac{2\lfloor x\rfloor }{x^3  }dx < \int_1^\infty \frac{2f(x)}{x^3} dx = \frac {10}{6}$$
Actually I am trying to show that $\large\frac {10}{6}$ is upper bound for $\zeta(2)$ using Euler Summation formula.
 A: $\displaystyle \int_1^{\infty} \frac{2\lfloor x \rfloor }{x^3}\ \mathrm{d}x =\sum_{n=1}^{\infty} \int_n^{n+1} \frac{2n}{x^3}\ \mathrm{d}x=\sum_{n=1}^{\infty}\frac{2 n+1}{n (n+1)^2}=\sum_{n=1}^{\infty} \frac{1}{n}-\frac{1}{n+1}+\frac{1}{(n+1)^2}=\frac{\pi^2}{6}$
to prove that : $\displaystyle \frac{\pi^2}{6} \leq \frac{5}{3} $ :
$S=\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} =\sum_{n=1}^{5}\frac{1}{n^2}+\sum_{n=6}^{\infty} \frac{1}{n^2} \leq \sum_{n=1}^{5}\frac{1}{n^2}+\sum_{n=6}^{\infty} \frac{1}{n(n-1)}$
$\displaystyle S=\sum_{n=1}^{5}\frac{1}{n^2}+\sum_{n=6}^{\infty} \frac{1}{n}-\frac{1}{n-1}=\frac{5989}{3600}\leq \frac{10}{6}$
A: If it's just about the sum of reciprocal squares use the following estimate: For any $r\geq1$ one has
$$\eqalign{\sum_{k=1}^\infty{1\over k^2}&< \sum_{k=1}^r{1\over k^2}+\sum_{k=r+1}^\infty{1\over k^2-{1\over4}}\cr &=\sum_{k=1}^r{1\over k^2}+\sum_{k=r+1}^\infty\left({1\over k-{1\over2}}-{1\over k+{1\over2}}\right)\cr
&=\sum_{k=1}^r{1\over k^2}+{1\over r+{1\over2}}\ .\cr}$$
Already $r:=1$ gives the desired ${5\over3}\doteq1.667$; with $r:=2$ you get $1.65$.
