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.

Hint: Note that
$$
\begin{align}
\frac1{(k-1)^3}-\frac1{k^3}
&=\frac{3k^2-3k+1}{k^3(k-1)^3}\\
&\gt\frac{3k^2-3k}{k^3(k-1)^3}\\
&=\frac3{k^2(k-1)^2}\\
&\gt\frac3{k^4}
\end{align}
$$
Therefore,
$$
\begin{align}
\sum_{k=n}^\infty\frac1{k^4}
&\lt\frac13\sum_{k=n}^\infty\left(\frac1{(k-1)^3}-\frac1{k^3}\right)\\
&=\frac1{3(n-1)^3}
\end{align}
$$

