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)=\frac1{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
`\frac1{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(\mathrm dx)/(x^3)`

`=\displaystyle\lim_(t\to\infty)\int_1^t(\mathrm dx)/(x^3)`

`=-2\displaystyle\lim_(t\to\infty)\frac1{x^2}\bigg|_(x=1)^(x=t)`

`=-2\lim_(t\to\infty)\left(\frac1{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.