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How can I simplify this expression:

$$\large \frac {2^n}{n^{nlog(2)}(log(2))^{nlog(2)}}$$

My guess is that somehow I can extract a $2^n$ factor from the denominator, but how can I do that?

Would it be a change of base for the logarithms? But would a change of base ... change the answer? E.g., say I were to show that the infinite sum of this expression converges, and the logarithms given are probably assumed to have base e. Can I just change the logarithm without changing the answer?

Sorry for the wordy and simple question, but this is a bit tricky for me. And the change of base technique has shown up a few times in past questions on advanced calculus, so I just want to be sure how to do it properly. I haven't found anything on this, other than some simple formulas for base change from various online sources. But I haven't found anything that addresses the issue of raising a logarithm to a power of a logarithm.

Thanks,

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    $\begingroup$ What makes you think there is a simplification? $\endgroup$ – Thomas Andrews Oct 17 '15 at 3:08
  • $\begingroup$ Good point @ThomasAndrews. I might have simplified enough at this point. But the $2^n$ on the numerator, I think, makes it problematic for convergence. Can I conclude from here that the infinite series of this summand is convergent? Hmm... $\endgroup$ – User001 Oct 17 '15 at 3:10
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    $\begingroup$ If you want to find out if the series is convergent, you can take the $n$th root of each component in the fraction. $\endgroup$ – Thomas Andrews Oct 17 '15 at 3:12
  • $\begingroup$ ...I wish I had saw that myself. I'll try not to forget the root test, going forward. Thanks so much, @ThomasAndrews :-) $\endgroup$ – User001 Oct 17 '15 at 3:30
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To reiterate, we have the expression $$\dfrac{2^n}{n^{n\log 2}(\log 2)^{n\log 2}}$$

which looks a bit messy.

If we assume the logarithms given have base $e$ and that $n$ is a positive integer, then we can use basic properties of exponents and logs to tidy it up some.

[reminder: I'm using the convention $\log x=\log_e x=\ln x$, not to be confused with the alternative notation $\log x=\log_{10} x$.]

There are several ways to rewrite the above expression more compactly. One might consider $$\left(\dfrac{e}{n\log 2}\right)^{n\log 2}$$ which is "prettier" and allows the root test to be applied more readily.

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  • $\begingroup$ Hi @user281097, thanks so much for your answer. It is indeed neater and it gives another example of using the e^log technique to rewrite functions and simplify, especially when evaluating limits. Have a great night :-) $\endgroup$ – User001 Oct 17 '15 at 7:03
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    $\begingroup$ @LebronJames np :) I assume you were able to show the series $$\displaystyle\sum_{n=1}^\infty \left(\dfrac{e}{n\log 2}\right)^{n\log 2}$$ is absolutely convergent. As an addendum, did you have any interest in computing the limit sum? I can't seem to find an easy way to do it. $\endgroup$ – Corellian Oct 21 '15 at 20:58
  • $\begingroup$ Hi @Brody - sorry for the late response. I have had some frustrating late nights of mathematics recently. The original series that I was looking at was $\large \sum_{n=2}^{\infty} \frac{1}{log(n)^{log(n)}}$. And, nope, I just had to show that it converges. It was a part(b) question. Part (a) was considerably easier; with the summand = $\large \frac{1}{(log(n))^3}$. The Cauchy condensation test was an obvious choice to apply to part(a), not so obvious for part(b) -- at least for me it wasn't. $\endgroup$ – User001 Oct 23 '15 at 5:32
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    $\begingroup$ If you read the comments above by Thomas Andrews, the condensation test was perfectly fine, but I had to realize that the summand was now put in a from where the root test can be applied. Your using the e^log technique was really cool - that of course shows up everywhere and I wish I would have saw that myself. Thanks again for your answer @Brody :-) $\endgroup$ – User001 Oct 23 '15 at 5:32
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This might be somewhat simpler: $$\large \frac {2^n}{(log(2^n))^{log(2^n)}}$$

I don't think that you could further simplify.

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  • $\begingroup$ Hi @Moti, interesting ... are you saying that this form is simpler to prove series convergence? In what way? My summand is the simplification of yours, actually, and it is crucial, it seems like, since it allowed me to use the root test, as Thomas Andrews had hinted in the comments. But I'm open to hearing your thoughts. Thanks, $\endgroup$ – User001 Oct 17 '15 at 5:31
  • $\begingroup$ Why not use the derivative of denominator ant numerator for $$\large \frac {x}{(log(x))^(log(x))}$$? $\endgroup$ – Moti Oct 17 '15 at 5:49
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    $\begingroup$ @LebronJames "Are you saying that this form is simpler to prove series convergence?" That would be a remarkable stretch given that you didn't mention anything about proving convergence being the purpose of simplification. The expression in the answer is aesthetically much simpler. $\endgroup$ – Erick Wong Oct 17 '15 at 5:51
  • $\begingroup$ Hi @ErickWong - sorry, I should have added more context to my question, but I agree that the above answer is in much simpler form. $\endgroup$ – User001 Oct 17 '15 at 6:42

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