Question

Convergence of riemann zeta function when s=3

Answer 1

$\zeta(s)=\displaystyle\sum_{n=1}^\infty\frac1{n^s}$
$\zeta(3)=\displaystyle\sum_{n=1}^\infty\frac1{n^3}$

Writing the infinite series as a limit of a partial sum, you can see that the series represents the left-endpoint Riemann approximation to the function $f(x)=\dfrac1{x^3}$:

$\displaystyle\sum_{n=1}^\infty\frac1{n^3}=\lim_{N\to\infty}\sum_{n=1}^N\frac1{n^3}=\int_1^\infty f(x)\,\mathrm dx$

In particular, since $\dfrac1{x^3}$ is a concave function for $x>0$, i.e. strictly decreasing and approaching 0 as $x\to\infty$, which means the Riemann sum approximation is greater than the value of the integral.

$\displaystyle\sum_{n=1}^\infty\frac1{n^3}\ge\int_1^\infty\dfrac{\mathrm dx}{x^3}$
$=\displaystyle\lim_{t\to\infty}\int_1^t\frac{\mathrm dx}{x^3}$
$=-2\displaystyle\lim_{t\to\infty}\frac1{x^2}\bigg|_{x=1}^{x=t}$
$=-2\lim_{t\to\infty}\left(\dfrac1{t^2}-1\right)$
$=2$

So the value of the series is some real number smaller than 2 (actually somewhere around 1.202).

Long story short: use the integral test for determining whether the series converges.