# For what values for m does $\sum \limits_{k=1}^{\infty}\frac{1}{(\ln{k})^m}$ converges?

For what values for m does $$\sum \limits_{k=2}^{\infty}\frac{1}{(\ln{k})^m}$$ converge? What about $$\sum_{k=2}^{\infty}\frac{1}{(\ln(\ln{k}))^m}$$ or more generally $$\sum_{k=2}^{\infty}\frac{1}{(\ln(\cdots (\ln{k}))\cdots)^m}$$ ?

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Better watch where you start your sums. $\log 1=0$. –  Gerry Myerson Jul 30 '12 at 8:45
thanks Gerry, done! –  Arjang Jul 30 '12 at 21:48
Not quite there yet - what's $\log\log\log2$? and what about the title? –  Gerry Myerson Jul 30 '12 at 23:11
I suggest defining $\ln^+ n = \max\{\ln n, 1\}$ and then replacing all $\ln$s with $\ln^+$s. –  JeffE Jul 31 '12 at 2:54

Let $\alpha,\beta \in \mathbb{R}$ $$\sum_{k=2}^\infty \frac{1}{k^\alpha \ln(k)^\beta}$$ is convergent iif $(\alpha>1)$ or $(\alpha=1, \ \beta>1)$

So in your case it is not convergent. As

$$\frac{\sqrt{k}}{\ln(k)^\beta} \rightarrow \infty$$ we can find $k_0$ such that for all $k \geq k_0$ $$\frac{\sqrt{k}}{\ln(k)^\beta} > 1$$ which yields $$\frac{1}{\ln(k)^\beta} > \frac{1}{\sqrt{k}}$$ The right-hand side behind divergent, your series is divergent. These series are called Bertrand series. See here for proofs.

As for your second series, just repeat the same argument. Take $\ln(k) > k_0$ you will get $$\frac{1}{\ln(\ln(k))^\beta} > \frac{1}{\sqrt{\ln(k)}}$$ The right-hand side was just proven to be divergent. Repeat the process for your generalized version.

Edit: You have to be careful though about the integer that will start the series. For your first series it is obvisouly $k=2$ because $\ln(1)=0$ which yields division by zero. For the second series the division by zero occurs at $e$ and $\ln(\ln(k))$ is not defined for $k=1$ so start at $k=3=\lfloor e \rfloor+1$. For you generalized series just start at $k=\lfloor e^{e^{\cdots^e}} \rfloor +1$

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I guess, the OP was asking for a proof... –  draks ... Jul 30 '12 at 8:51
Just added the proof. –  vanna Jul 30 '12 at 8:59

For the first series,we can use Cauchy's condensation criterion when $p>0$ (when $p\leq 0$ the series is divergent):

If $\{a_n\}$ is a decreasing sequence of positive numbers, the series $\sum_ka_k$ is convergent if and only if $\sum_k2^ka_{2^k}$ is convergent.

We have with $a_k=\frac 1{(\ln k)^m}$ that $2^ka_k=\frac{2^k}{k^m(\ln 2)^m}$ and it diverges.

An alternative way is to note that $\ln k\leq k^{1/m}$ for a $m$ large enough (which depend on $m$), and use the divergence of harmonic series.

We have for $x\geq 1$ that $\log x\leq x$. Denote $f_N$ the $N$-th iterate of the logarithm. For a fixed $N$, we can find $n(N)$ such that for $k\geq n(N)$, $f_{N-1}(k)\geq 1$. Hence $f_N(k)\leq \log k$ for $k$ large enough, which shows that the series is divergent for all $m$ and all $N$.

The problem here is that the logarithm grows too slowly. Taking iterates doesn't make the thing better, and the exponent doesn't change anything.

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