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, $$ \begin{align} \sum_{k=n}^\infty\frac1{k^4} &\lt\frac13\sum_{k=n}^\infty\left(\frac1{(k-1)^3}-\frac1{k^3}\right)\\ &=\frac1{3(n-1)^3} \end{align} $$
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 $$ 1\le S\le 1+\frac{1}{3}$$
