Respuesta :

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[tex]\zeta(s)=\displaystyle\sum_{n=1}^\infty\frac1{n^s}[/tex]
[tex]\zeta(3)=\displaystyle\sum_{n=1}^\infty\frac1{n^3}[/tex]

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 [tex]f(x)=\dfrac1{x^3}[/tex]:

[tex]\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[/tex]

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

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

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.

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