Hello;
What is the limit of $\displaystyle \frac{({\log{2}^n})^{\log{2}^n}}{({\log{2}^{n+1})} ^{\log{2}^{n+1}}}$ as $\displaystyle n$ tends to infinity? To my mind, it's either $\displaystyle \frac{1}{e}$ or $\displaystyle 0$. And I'm leaning towards $\displaystyle \frac{1}{e}$, but I'm not sure.
Thank you very much.
I assume you are using base 10 in your logarithms. The limit is zero.
This simplifies to:
$\displaystyle \left( {\frac {n}{n+1}} \right) ^{n\ln \left( 2 \right) } \left( {
\frac {1}{ \left( n+1 \right) \ln \left( 2 \right) }} \right) ^{\ln
\left( 2 \right) }$
If you are using base e. A similar simplification will come out of base 10. So the limit is a little more obvious.
$\displaystyle
\frac{({\log{2}^n})^{\log{2}^n}}{({\log{2}^{n+1})} ^{\log{2}^{n+1}}}
$ comes from the ratio test (all multiplied by $\displaystyle 2$) as an attempt to prove that $\displaystyle
\sum_{k=1}^{\infty}\frac{2^k}{({\log{2}^{k})}^{\lo g{2}^{k}}}
$, which is derived from $\displaystyle
\sum_{n=2}^{\infty}\frac{1}{({\log{n})}^{\log{n}}}$ (by the Cauchy condensation test), is convergent.
By the way, what is the best way to show that $\displaystyle
\sum_{n=2}^{\infty}\frac{1}{({\log{n})}^{\log{n}}}$ 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?Originally Posted by disclaimer
By the way, what is the best way to show that $\displaystyle
\sum_{n=2}^{\infty}\frac{1}{({\log{n})}^{\log{n}}}$ is convergent? Thanks.
If you apply the condensation test with e in place of 2, it tells you that the convergence of $\displaystyle \sum_{n=2}^{\infty}\frac{1}{({\log{n})}^{\log{n}}}$ is equivalent to that of $\displaystyle \sum\left(\frac en\right)^n$, which converges by comparison with $\displaystyle \sum\left(\frac 12\right)^n$.
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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