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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.2012-06-18

4 Answers 4

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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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    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.2012-06-18
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Hint: $n^{\frac{n}{\log(n)}}=e^n$

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

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    @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 $2012-06-18
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@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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    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 $e$.2012-06-18
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METHOD I

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.

METHOD II

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 oenamen. This way is even more elementary.2012-06-18