While working on problems from Spivak's Calculus, I came on one asking for the convergence/divergence of the series $$\sum_{n=1}^{\infty} \frac{1}{n^{1+\frac{1}{n}}}.$$

This is a straightforward comparison with the harmonic series to show divergence, but I ended up playing around with the idea a bit more and stumbling on the interesting fact that $$\sum_{n=2}^{\infty} \frac{1}{n^{1+\frac{1}{\log n}}}\,\,\text{diverges}\quad\text{and}\quad \sum_{n=2}^{\infty} \frac{1}{n^{1+\frac{1}{\log\log n}}}\,\,\text{converges}.$$

The first is another limit comparison with the harmonic series; the second I tried to do by looking at $$\int_2^{\infty} \frac{1}{n}\cdot \frac{1}{n^{\frac{1}{\log \log n}}}\, dn$$ and substituting $u=\log n$ to give $$\int_2^{\infty}e^{-\frac{u}{\log u}}\, du,$$ which seems more tractable.

I have two questions:

  1. Can anyone show convergence of divergence of the second series by finishing up my attempt or through some other technique?
  2. More generally, for the series $$\sum_{n=1}^{\infty} \frac{1}{n^{1+\frac{1}{f(n)}}}$$ what can be said about the growth rates/other characterizations of the $f$ for which this converges?

Any help would be much appreciated!


Since $\frac{u^{1/2}}{\log u}\to\infty$, there exists $c>0$ such that $\frac{u^{1/2}}{\log u}>1$ for all $u>c$. Thus $\frac{u}{\log u}>u^{1/2}$ for all $u>c$, and then $$ e^{-\frac u{\log u}}\leq e^{-u^{1/2}},\ \ u>c. $$ Also, as $\frac{e^{u^{1/2}}}{u^2}\to\infty$, we can find $d>c$ with $$e^{-u^{1/2}}\leq\frac1{u^2},\ \ u>d. $$

About your second question, you get the integral $$ \int_1^\infty\frac1x\,e^{-\frac{\log x}{f(x)}}\,dx, $$ so basically you need $f$ to lose against the logarithm so that the exponential retains some kick.


There was a request for a different approach from that discussed in the OP. Here, we use Cauchy's Condensation test on the series $S$ expressed as

$$S=\sum_{n=3}^\infty f(n) \tag 1$$

where the summand $f(n)$ is given by


Now, we analyze the sum $T$ expressed as

$$\begin{align} T&=\sum_{n=3}^\infty 2^n f(2^n)\\\\ &=\sum_{n=3}^\infty 2^n \frac{1}{2^{n\left(1+\frac{1}{\log(\log(2^n)}\right)}}\\\\ &=\sum_{n=3}^\infty 2^{-\left(\frac{n}{\log(n\log(2))}\right)} \tag 2 \end{align}$$

Now, for any $\alpha >0$, the logarithm function satisfies the inequality SEE THIS ANSWER

$$\log(x)\le \frac{x^\alpha -1}{\alpha}\le \frac{x^\alpha}{\alpha} \tag 3$$

Applying $(3)$ to $(2)$ reveals that for $0<\alpha <1$

$$2^{-\left(\frac{n}{\log(n\log(2))}\right)}\le 2^{-\left(\frac{\alpha n^{1-\alpha}}{\log^\alpha(2)}\right)}$$

Thus, the series in $(2)$ clearly converges by the comparison test. Take for example $\alpha =1/2$. Then, it is easy to show that

$$2^{-\left(\frac{ n^{1/2}}{2\sqrt{\log(2)}}\right)}\le \frac{48/\log^{3/2}}{n^{3/2}}$$

and the series converges by comparing it to the series $\sum_{n=3}^\infty \frac{1}{n^{3/2}}$.

Finally, the condensation test guarantees that the series in $(1)$ also converges. And we are done!


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