Simplest proof that $\zeta(s) \to \infty$ as $s \to 1$?

Question

For homework I had to prove the divergence of the series $1/(k\log^p k)$ for all real $p$ (it is simple to do so via integration.) However a more elegant means would be to appeal to the behavior of the zeta function $\zeta(s)=\sum\limits_{k=1}^\infty 1/k^s$ on the real line. What is the most elementary proof that $\lim\limits_{s\to 1}\zeta(s)=\infty$?

Answer 1

Your series does not diverge for all real $p$. By integration, you can show that in fact it converges if $p>1$ and diverges if $p<1$. Or else, if you want to avoid integration, you can use the Cauchy Condensation Test.

So the behaviour of the zeta-function is not directly relevant to your convergence problem.

As to proving that the limit of $\zeta(s)$ as $s$ approaches $1$ from the right is infinite, one way of doing it is to use integration to find a good lower bound for the sum in terms of $s$, and then let $s\to 1^+$.

Answer 2

For every $k\geqslant1$ and every $x\geqslant k$, $\dfrac1{k^s}\geqslant\dfrac1{x^s}$, hence $\displaystyle\frac1{k^s}\geqslant\int_k^{k+1}\frac{\mathrm dx}{x^s}$. Summing this over $k\geqslant1$, one gets $\zeta(s)\geqslant\displaystyle\int_1^{+\infty}\frac{\mathrm dx}{x^s}=\frac1{s-1}$. Since $\dfrac1{s-1}\to+\infty$ when $s\to1^+$, $\zeta(s)\to+\infty$.

Answer 3

It seems that the most elegant proof is Euler's proof via comparison \begin{align*} \zeta (1) &= 1 + \left( {\frac{1}{2}} \right) + \left( {\frac{1}{3} + \frac{1}{4}} \right) + \left( {\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}} \right) + \cdots \\ &> 1 + \left( {\frac{1}{2}} \right) + \left( {\frac{1}{4} + \frac{1}{4}} \right) + \left( {\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}} \right) + \cdots \\ &= 1 + \left( {\frac{1}{2}} \right) + \left( {\frac{1}{2}} \right) + \left( {\frac{1}{2}} \right) + \cdots \\ &= \infty. \end{align*}