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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 $\lim\limits_{s\to 1}\zeta(s)=\infty$?

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    $\begingroup$ To answer the title question: For real $s>1$, zeta is positive and monotonic, so limit then follows since the harmonic series diverges. An elementary way to prove the harmonic series diverges is by grouping the terms into chunks of size $2^k$ we get $$1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right) +\cdots$$ and the terms in each pair of parentheses sum to something greater then $\frac{1}{2}$. This means the series diverges, because we would be adding $\frac{1}{2}$ to itself an infinite number of times. $\endgroup$ Jan 29, 2012 at 20:31
  • $\begingroup$ @Eric Naslund: Does this work without a little more calculation? It is possible to imagine that the limit as $s\to 1^+$ is finite even though $\zeta(1)$ is not. $\endgroup$ Jan 29, 2012 at 20:38
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    $\begingroup$ @AndréNicolas: No, because it is monotonic. $\endgroup$ Jan 30, 2012 at 6:18
  • $\begingroup$ I think monotony is not enough... The function $f(s)=2-s$ for $s \in (1, \infty]$ and $f(1)=\infty$ is monotonic and $f(1)=\infty$ but $\lim_{s \to 1^{+}}f(s)=1$. $\endgroup$
    – N. S.
    Feb 15, 2012 at 23:29
  • $\begingroup$ @N.S.: Dear N.S., But in this case each term in the series at $s = 1$ is the limit of the corresponding term for $s > 1$, and since all terms in sight are positive, it follows that $\zeta(s) \to \zeta(1)$ as $s \to 1$. Regards, $\endgroup$
    – Matt E
    Oct 22, 2012 at 0:33

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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 $s\to1^+$, $\zeta(s)\to+\infty$.

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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 $s\to 1^+$.

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    $\begingroup$ You mean $s \to 1^+$, right? $\endgroup$
    – cardinal
    Jan 29, 2012 at 20:24
  • $\begingroup$ @cardinal: Thanks, I certainly do! Fixed. $\endgroup$ Jan 29, 2012 at 20:28
  • $\begingroup$ Either proof method mentioned in the first paragraph also shows that the series diverges if $p=1$. $\endgroup$ Feb 15, 2012 at 17:02
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    $\begingroup$ Dear Andre, As you no doubt know, the series for integral $p \leq 0$ is obtained via a $|p|$-fold differentiation of $\zeta(s)$, and then setting $s = 1$. So if one knows e.g. that $\zeta(s)$ has a first order pole at $s = 1$, one obtains the divergence for integral $p \leq 0$, and then for real $p \leq 0$ by monotonicity. So it is not completely fair to dismiss the relationship to the $\zeta$-function. Best wishes, $\endgroup$
    – Matt E
    Oct 22, 2012 at 0:38
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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*}

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