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The series

$$\sum_{n=1}^{\infty} \frac{1}{n^{{n}/{\log(n)}}}$$

converges according to Wolframalpha.

Now I am not sure what the best technique is handling this one. I am thinking about a comparison test.

Here is what I thought $n \geq 1 \iff \log(n) \geq 1 \iff \dfrac{n}{\log(n)} \geq 1 \iff n^{\dfrac{n}{\log(n)}} \geq 1 \iff 0 \leq \frac{1}{n^{\frac{n}{\log(n)}}} \leq 1 \iff \sum_{n=1}^{\infty} 0 \leq \sum_{n=1}^{\infty} \frac{1}{n^{\frac{n}{\log(n)}}} \leq \sum_{n=1}^{\infty} 1 $

So by the Comparison Test, it converges. Or I guess i "sqqqqququuuuzed" the sum =)

Now my concern is that my sum is bounded, but I guess that doesn't imply the sum exist because something like $\sin(n)$ diverges even though it is bounded. Any insights?

EDIT: $\log(n)$ isn't the natural log

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Since all terms are positive it's monotonically increasing, and so showing it's bounded implies that the sum exists. – Robert Mastragostino Jun 18 '12 at 0:06
up vote 13 down vote accepted

When $n>1$, $n^{\frac{n}{\log n}}=e^{\frac{n}{\log n}\log n}=e^n$.

Then it's a geometric series : $$\sum_{n=2}^\infty\left(\frac{1}{e}\right)^n$$ with $\frac{1}{e}<1$.

So the series conveges.

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Perhaps, when $n>1$. – robjohn Jun 17 '12 at 23:44
Indeed, corrected. And you were faster. – JBC Jun 17 '12 at 23:45
So I am wrong? What's wrong with my proof? Also I am confused with $n^{\frac{n}{\log n}}=e^{\frac{n}{\log n}\log n}$. How did you change n = e? – Hawk Jun 18 '12 at 0:21
I don't really understand your proof... For example $$\sum_{n=0}^\infty1=\infty$$ So you didn't show that the sum is bounded. And your equivalences seem wrong to me. I noticed that $n^{\frac{n}{\log n}}=e^n$, so $\frac{1}{n^\frac{n}{\log n}}=\frac{1}{e^n}=\left(\frac{1}{e}\right)^n$. – JBC Jun 18 '12 at 0:25
Indeed. However, if the first (or one) term isn't finite/well definite, the exercice isn't interesting. So removing it seems a good solution ;-p. – JBC Jun 18 '12 at 1:59

Hint: $n^{\frac{n}{\log(n)}}=e^n$

Although the $n=1$ term will have to be handled with care :-)

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Where did you get $n^{\frac{n}{\log(n)}}=e^n$ from? – Hawk Jun 18 '12 at 0:37
@Jak, it is a basic equality: $$a^b=e^{b\log a}\,\,,\,\,a>0\,,\,\log=\,\text{ the natural logarithm, as it's customary unless otherwise stated}\,$$ You can prove easily the equality using the basic property of logarithms:$$\log x^b=b\log x$$which is true for logarithms to any base. – DonAntonio Jun 18 '12 at 2:12
@jak: $n=e^{\log(n)}$, so raising both sides to the $\frac{n}{\log(n)}$ power yields $$ n^{\frac{n}{\log(n)}}=e^{\log(n)\frac{n}{\log(n)}}=e^n $$ – robjohn Jun 18 '12 at 2:14

@JBC and @robjohn dealt nicely with base $e$.

In base $b$, $n^{n/\log_b(n)} = b^n$, since $n=b^{\log_b n}$. (That is, the exponential function is the inverse function of the logarithmic function.) In fact, $\lim_{n\to 1} n^{n/\log_b(n)} = b$, so the formula can be used for all relevant $n$. Thus, $$\begin{eqnarray*} \sum_{n=1}^\infty \frac{1}{n^{n/\log_b(n)}} &=& \sum_{n=1}^\infty \left(\frac{1}{b}\right)^n. \end{eqnarray*}$$ This is just a geometric series which converges to $\frac{1}{b-1}$ for $b>1$.

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Actually how did you assume log(n) is e-based? – Hawk Jun 18 '12 at 1:57
@jak: I did not assume it. I can see JBC and robjohn did since $e$ appeared as the base of the exponential function. Stare at $n=b^{\log_b n}$ for a bit. – user26872 Jun 18 '12 at 2:04
@jak By convention, logs that are given without any base are more often than not taken to be natural logs in analysis. Occasionally in computer science and analysis of algorithms 'log' will mean log base 2, but the confusion shows up often enough that computer scientists have taken to using 'lg' for $\log_2$. (I believe this started with Don Knuth, but I can't say with any certainty.) 'Log' for $\log_{10}$ is seldom used, since natural logs show up so much more often. – Steven Stadnicki Jun 18 '12 at 2:11
@StevenStadnicki: To add another wrinkle, physical scientists often write $\ln x$ for the natural log and $\log x$ for the common log. – user26872 Jun 18 '12 at 2:17
In French we usually write $\mathrm{ln}$ in base $e$ and $\log$ in base $10$. And I think that in English we generally write $\log$ in base $e$, and sometimes $\mathrm{Log}$ in base $10$. Since it's not my native language, I may be wrong. But, it's why I considered your $\log$ in base $e$. – JBC Jun 18 '12 at 2:25


It's easy to see that:

$$\sum_{n=1}^{\infty} \frac{1}{n^{\frac{n}{\log(n)}}}\leq\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}\leq 1+\sum_{n=2}^{\infty} \frac{1}{n(n-1)}=2$$

We may conclud that the sum converges.


Cauchy condensation test works pretty fast, as well.

Q.E.D. (these are just 2 alternative ways to the geometric series)

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Hi Chris. While the first method works for base $e$, if the base of the log is $b$, and $1<b<1+6/\pi^2$, the sum is actually larger than $\pi^2/6$. – user26872 Jun 18 '12 at 9:04
Hi oenamen. This way is even more elementary. – user 1618033 Jun 18 '12 at 9:24

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