I'm trying to find out now whether the series $\sum_{n=2}^{\infty } a_{n}$ converges or not when $$a_n = \frac{n^{\log n}}{(\log n)^{n}}$$

Again, I tried d'Alembert $\frac{a_{n+1}}{a_{n}}$, Cauchy condensation test $\sum \limits_{n=2}^{\infty } 2^{n}a_{2^n}$, and they both didn't work for me.

I can't use Stirling, nor the integral test.

Edit: I'm searching for a solution which uses sequences theorems and doesn't involve functions.

Thank you

  • 2
    $\begingroup$ Do you mean $(\log n)^n$ or $\log(n^n)$? If you mean the first one, it appears $a_n \to 0$. If you mean the second, it appears the terms go to $\infty$. $\endgroup$ – Graphth Apr 12 '11 at 15:11
  • $\begingroup$ Thank Numth for the comment, I edited that. $\endgroup$ – user6163 Apr 12 '11 at 15:12
  • $\begingroup$ So, do you mean $\log(n^n)$? Note that $\log(n^n) = n\log(n)$. $\endgroup$ – Arturo Magidin Apr 12 '11 at 15:17
  • $\begingroup$ No, this is my final Edit, I am sorry :-) $\endgroup$ – user6163 Apr 12 '11 at 15:19

A comparison test will work here; the key is to write both numerator and denominator in terms of exponentials with bases not involving $n$. Note that the numerator is $e^{\log^2 n}$, which is less than $e^{n/2}$ for sufficiently large $n$. The denominator is $e^{n \log \log n}$, which is greater than $e^n$ for sufficiently large $n$; so for all sufficiently large $n$ the terms are less than $e^{-n/2}$ and thus the series converges.


Assuming he means $\log(n^n)$:

Intuitively, $\log n \leq n$ for all $n$ greater some $n_0$. So picking an $n_0$ such that $\log n \leq n$ and $(\log n) - 1 \geq 1$ will yield $\frac{n^{(log n)-1}}{\log n} \geq 1$. So the series diverges.


Note that $n^{\log(n)}=e^{\log(n)^2}$ and $\log(n)^n=e^{n\log(\log(n))}$. Thus $$a_n=e^{\log(n)^2-n\log(\log(n))},$$ and $a_n\rightarrow0$ iff $\log(n)^2-n\log(\log(n))\rightarrow-\infty$ (which I'm pretty sure it does). If you can find a sequence which bounds $\log(n)^2-n\log(\log(n))$ from above and also goes to $-\infty$ fast enough, you should be able to prove that the sum converges.

(this is for the previous version with $\log(n^n)$ instead of $\log(n)^n$)

If $a_n\not\rightarrow0$, then the series $\sum a_n$ must diverge. Note that $a_n=f(n)$ where $$f(x)=\frac{x^{\log(x)}}{\log(x^x)}=\frac{x^{\log(x)}}{x\log(x)}=\frac{x^{\log(x)-1}}{\log(x)}.$$ Thus, if we show that $\lim_{x\rightarrow\infty}f(x)\neq0$, then the series must diverge.

We have that $$\lim_{x\rightarrow\infty}\frac{x^{\log(x)-1}}{\log(x)}=\frac{\infty}{\infty}$$ so using L'Hopital this equals $$\lim_{x\rightarrow\infty}\frac{(\log(x)-1)x^{\log(x)-2}\cdot\frac{1}{x}}{\frac{1}{x}}=\lim_{x\rightarrow\infty}(\log(x)-1)x^{\log(x)-2}=\infty$$


I used Cauchy condensation, then comparison test, then root test and it seems to converge:

Condensation test:

$$\sum \limits_{n=2}^{\infty }2^n\frac{2^{n^{n\log 2}}}{(n\log 2)^{2^{n}}}$$

Comparison test:

$$\sum \limits_{n=2}^{\infty }2^n\dfrac{2^{n^{n\log 2}}}{(n\log 2)^{2^{n}}}< \sum \limits_{n=2}^{\infty }2^n\dfrac{n^{n}}{n^{2^{n}}}$$

Root test:

$$\sqrt[n]{\frac{2^nn^{n}}{n^{2^{n}}}} \underbrace{\longrightarrow}_{n \to \infty}0$$

Then the series converges


root test works fine:

$$\limsup_{n\to\infty} \sqrt[n] {\frac{n^{\log n}}{(\log n)^n}}=\limsup_{n\to\infty} \frac{\sqrt[n] {n^{\log n}}}{\sqrt[n] {(\log n)^n}}=\limsup_{n\to\infty}\frac{1}{\log n}$$

and for all $n>$(the base of the logarithm): $$\frac{1}{\log n}<1$$ therefor this series converges by the root test (and we are able to use this test because $a_n \geq0$) .


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy