Using integral test for convergence of $\sum_{n = 1}^{\infty} \frac 1 {n^3}$ How do I show that the series 
$\:\large\sum_\limits{n=1}^{\infty}\large{\frac 1 {\:n^3}}\:$
is convergent using the integral test ?
 A: An alternative to the integral test is the technique called "creative telescoping":
$$\begin{eqnarray*} \color{red}{\zeta(3)} = 1+\sum_{n\geq 2}\frac{1}{n^3}&\color{red}{\leq}& 1+\sum_{n\geq 2}\frac{1}{(n-1)n(n+1)}\\&=&1+\frac{1}{2}\sum_{n\geq 2}\left(\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1}\right)\\&=&1+\frac{1}{2}\sum_{n\geq 2}\left(\frac{1}{n-1}-\frac{1}{n}\right)-\frac{1}{2}\sum_{n\geq 2}\left(\frac{1}{n}-\frac{1}{n+1}\right)\\&=&1+\frac{1}{2}-\frac{1}{4}=\color{red}{\frac{5}{4}}.\end{eqnarray*} $$
The bound is quite tight and the technique can be generalized in order to prove Apery's identity:
$$ \zeta(3) = \frac{5}{2}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^3\binom{2n}{n}}. $$
A: $$
\sum_{n=1}^\infty \frac 1 {n^3}
= 1 + \sum_{n=2}^\infty \frac 1 {n^3}
\le 1 + \sum_{n=2}^\infty \int_{n-1}^n \frac 1 {x^3}\,dx
= 1 + \int_1^\infty \frac 1 {x^3} \, dx = 1 + \frac 1 2.  
$$
A: By Integral test,
we have $$\int_N^\infty \frac{1}{n^3} dn = \left[-\frac{1}{2n^2}\right]_N^\infty=\frac{1}{2N^2}$$ and this improper integral quite evidently exists. Hence the series is convergent.
A: Here $f(n)=\frac{1}{n^3}$ so $f(x)=\frac{1}{x^3}$. Then for $N\in \mathbb N$ $$\int_{1}^{N}\frac{1}{x^3}\,dx=\left[-\frac{1}{2x^2}\right]^{N}_1=\frac12-\frac{1}{2N^2}\to \frac12$$ as $N\to \infty$.
