Proof that $\frac{9}{8} < \sum\limits_{n=1}^{\infty} \frac{1}{n^3} < \frac{5}{4}$.

Approach Avoiding Integrals

Note that $$ \begin{align} \frac1{\left(n-\frac12\right)^2}-\frac1{\left(n+\frac12\right)^2} &=\frac{2n}{\left(n^2-\frac14\right)^2}\\ &\gt\frac2{n^3} \end{align} $$ Therefore, $$ \begin{align} \sum_{n=1}^\infty\frac1{n^3} &\lt1+\frac12\sum_{n=2}^\infty\left[\frac1{\left(n-\frac12\right)^2}-\frac1{\left(n+\frac12\right)^2}\right]\\ &=\frac{11}9 \end{align} $$ The other direction is simply $$ \begin{align} \sum_{n=1}^\infty\frac1{n^3} &\gt\sum_{n=1}^2\frac1{n^3}\\ &=\frac98 \end{align} $$ Thus, we get the tighter bounds $$ \frac98\lt\sum_{n=1}^\infty\frac1{n^3}\lt\frac{11}9 $$

---

Bounding by Integrals

Note that $$ \int_n^{n+1}\frac1{x^3}\,\mathrm{d}x\le\frac1{n^3}\le\int_{n-1}^n\frac1{x^3}\,\mathrm{d}x $$ Therefore, $$ 1+\overbrace{\int_2^\infty\frac1{x^3}\,\mathrm{d}x}^{\le\sum\limits_{k=2}^\infty\frac1{n^3}}\le\sum_{n=1}^\infty\frac1{n^3}\le1+\frac18+\overbrace{\int_2^\infty\frac1{x^3}\,\mathrm{d}x}^{\ge\sum\limits_{k=3}^\infty\frac1{n^3}} $$ Thus, $$ \frac98\le\sum_{n=1}^\infty\frac1{n^3}\le\frac54 $$