Simplest proof that $\zeta(s) \to \infty$ as $s \to 1$? 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 $\lim\limits_{s\to 1}\zeta(s)=\infty$?
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 $s\to1^+$, $\zeta(s)\to+\infty$.
It seems that the most elegant proof is Euler's proof via comparison
\begin{align*}
\zeta (1) &amp;= 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 \\
&amp;&gt; 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 \\
&amp;= 1 + \left( {\frac{1}{2}} \right) + \left( {\frac{1}{2}} \right) + \left( {\frac{1}{2}} \right) + \cdots \\
&amp;= \infty.
\end{align*}