limit

**disclaimer** · Mar 7th 2008, 02:23 PM

Hello;

What is the limit of $\frac{({\log{2}^n})^{\log{2}^n}}{({\log{2}^{n+1})} ^{\log{2}^{n+1}}}$ as $n$ tends to infinity? To my mind, it's either $\frac{1}{e}$ or $0$ . And I'm leaning towards $\frac{1}{e}$ , but I'm not sure.

Thank you very much.

**spammanon** · Mar 7th 2008, 05:33 PM

I assume you are using base 10 in your logarithms. The limit is zero.
This simplifies to:

$\left( {\frac {n}{n+1}} \right) ^{n\ln \left( 2 \right) } \left( { \frac {1}{ \left( n+1 \right) \ln \left( 2 \right) }} \right) ^{\ln \left( 2 \right) }$

If you are using base e. A similar simplification will come out of base 10. So the limit is a little more obvious.

**disclaimer** · Mar 7th 2008, 05:54 PM

Yeah..... I just found that out myself.... Thanks for your help anyway

**disclaimer** · Mar 7th 2008, 09:06 PM

$ \frac{({\log{2}^n})^{\log{2}^n}}{({\log{2}^{n+1})} ^{\log{2}^{n+1}}} $ comes from the ratio test (all multiplied by $2$ ) as an attempt to prove that $ \sum_{k=1}^{\infty}\frac{2^k}{({\log{2}^{k})}^{\lo g{2}^{k}}} $ , which is derived from $ \sum_{n=2}^{\infty}\frac{1}{({\log{n})}^{\log{n}}}$ (by the Cauchy condensation test), is convergent.

By the way, what is the best way to show that $ \sum_{n=2}^{\infty}\frac{1}{({\log{n})}^{\log{n}}}$ is convergent? Thanks.

**Opalg** · Mar 8th 2008, 07:15 AM

Originally Posted by disclaimer

By the way, what is the best way to show that $ \sum_{n=2}^{\infty}\frac{1}{({\log{n})}^{\log{n}}}$ is convergent? Thanks.

How about using the extended version of the Cauchy condensation test, which says that you can replace the number 2 by any real number greater than 1?

If you apply the condensation test with e in place of 2, it tells you that the convergence of $\sum_{n=2}^{\infty}\frac{1}{({\log{n})}^{\log{n}}}$ is equivalent to that of $\sum\left(\frac en\right)^n$ , which converges by comparison with $\sum\left(\frac 12\right)^n$ .

Of course, if the logs are to base 10 then the same argument will work. You just replace 2 by 10 instead of e.

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