Convergence criteria for real Riemann zeta function

Question

It is well known that the following series converges for $x > 1$, where $x$ is real.

$$ \zeta(x) = \sum_{n=1}^{\infty} \frac{1}{n^x}$$

Why is this? I have searched and can't find an answer, and most sources read as if it is so obvious it doesn't need an explanation.

For $x=1$ I know this is the divergent harmonic series.

A ratio test fails to conclude convergence:

$$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{(n+1)^x}{n^x} = 1$$

I've even plotted a graph of this and observed that as $x$ varies between 0 and 10, the limit still approaches 1.

I'm not mathematically trained so I'd appreciate explanations with minimum technical language.

Answer 1

We have that

$$\sum_{n=1}^{\infty} \frac{1}{n^x}\le1+\int_1^{\infty}\frac{1}{t^x}\,dt$$

and

$$\int_1^{a}\frac{1}{t^x}\,dt=\left[\frac1{1-x}t^{1-x}\right]_1^{a}=\frac1{1-x}a^{1-x}-\frac1{1-x}$$

thus for $x>1$

$$\int_1^{\infty}\frac{1}{t^x}\,dt=\lim_{a\to \infty}\int_1^{a}\frac{1}{t^x}\,dt=\frac1{x-1}$$

Answer 2

I thought it might be instructive to present an approach that relies on neither the integral test nor the Cauchy condensation test. Rather we make use of creative telescoping. To that end we proceed.

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First we note that for $x>1$

$$\frac{1}{n^x}=\frac{n^{1-x}-(n-1)^{1-x}}{1-x}+O\left(\frac{1}{n^{1+x}}\right)\tag1$$

We can show that $\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}{6}$ by a host of methodologies. And hence, for any $\epsilon>0$, the series $\sum_{n=1}^\infty \frac{1}{n^{2+\epsilon}}$ converges.

Therefore, summing both sides of $(1)$ and exploiting the telescoping series reveals

$$\sum_{n=1}^N \frac1{n^x}=1+\frac{1-N^{1-x}}{x-1}+\sum_{n=2}^N O\left(\frac{1}{n^{1+x}}\right)\tag2$$

For $x>1$, the second term on the right-hand side of $(1)$ approaches $\frac1{x-1}$ as $N\to \infty$ while the second term converges faster than $\sum_{n=1}^\infty \frac{1}{n^2}$.

Hence, the series $\sum_{n=1}^\infty \frac{1}{n^x}$ converges.

Answer 3

I don't know why you think that Cauchy Condensation involves complex analysis.

CC Suppose $a_n\ge0$ and $a_{n+1}\le a_n$. The series $\sum a_n$ converges if and only if the series $\sum 2^na_{2^n}$ converges.

No complex numbers in sight.

Now suppose $x>0$ and $a_n=1/n^x$. Then $$2^na_{2^n}=2^{(1-x)n},$$and it's clear that $\sum 2^{(1-x)n}$ converges if and only if $1-x<0$.