If $S=1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}\cdots$, then what is $\lfloor S \rfloor$? 
If 
  $$S=1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}\cdots$$ 
  then $$\lfloor S \rfloor  = \text{?}$$

What I tried: 
I know that 
$$S=1+\frac{1}{2^4}+\frac{1}{3^4}+\cdots=\zeta(4)=\frac{\pi^4}{90}\approx 1.1$$
then $\lfloor S \rfloor =1$.
But how do I find with inequality? Please have a look.

Note that $$\frac{1}{k^4}=\int_{k-1}^{k}\frac{1}{k^4}\,dx\le \int_{k-1}^k\frac{1}{x^4}\,dx$$
Therefore, $$\sum_{k=2}^n \frac{1}{k^4}\le \sum_{k=2}^n\int_{k-1}^k\frac{1}{x^4}\,dx=\int_1^n \frac{1}{x^4}\,dx=\frac{1}{3}\left[1-\frac{1}{n^3}\right]$$
Taking limit as $n\to\infty$, we get $$\sum\limits_{k=2}^\infty \frac{1}{k^4}\le \frac{1}{3}$$
Hence we can say that $$ 1\le S\le 1+\frac{1}{3}$$
