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I am trying to show for which parameter $\alpha$ does the sum$$\sum_{n=1}^{\infty} (n^{n^\alpha}-1)$$ converge. However I dont really know which criteria use.

mr.pink
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1 Answers1

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Since $\mathrm e^x\ge1+x$ for all $x$, we have

$$ n^{n^\alpha}-1=\mathrm e^{n^\alpha\log n}-1\ge n^\alpha\log n\;. $$

Since the terms are non-negative, the series can only converge if the series $\sum_nn^\alpha\log n$ converges. This is the case when $\alpha\lt-1$.

Conversely, let $\alpha\lt-1$. Then

$$ n^{n^\alpha}-1=\mathrm e^{n^\alpha\log n}-1=O\left(n^\alpha\log n\right)\quad\text{for $n\to\infty$}\;, $$

and the series converges.

joriki
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  • thanks, this makes sence however how do I know that he series $\sum_{n}^{} n^\alpha log (n)$ converges for $\alpha <-1$ ? – mr.pink Apr 04 '20 at 19:10
  • @mr.pink: You can show this e.g. using the integral test, or with the comparison test, comparing with $n^{\frac{\alpha-1}2}$. – joriki Apr 04 '20 at 19:15
  • I came back to this and I dont seem to understand how does a smaller sum ($\sum_{n} n^{\alpha} \log n$) determine the convergence of a larger sum .. am I missing something ? – mr.pink Apr 05 '20 at 14:59
  • @mr.pink: I'm sorry, I don't know which part of the argument you're referring to there. Generally speaking, a smaller sum can indeed be used to prove the convergence of a larger sum; for instance, the fact that $\sum_n\frac1{n^2}$ converges can be used to show that $\sum_n\frac2{n^3}$ converges, despite this sum being larger. The value of a series depends on its initial values, whereas its convergence doesn't. But if you point to where in the argument you were referring to, I might be able to say more. – joriki Apr 05 '20 at 15:26
  • I will try to explain myself: The part I understand is that $n^{n^\alpha}-1=\mathrm e^{n^\alpha\log n}-1\ge n^\alpha\log n $ and I also understand the final conclusion of $n^{n^\alpha}-1=\mathrm e^{n^\alpha\log n}-1=O\left(n^\alpha\log n\right)\quad\text{for $n\to\infty$} $ My problem is why does the convergence of series $\sum_nn^\alpha\log n$ mean the convergence of the original series. Isnt $\sum_nn^\alpha\log n$ smaller than $\sum_{n=1}^{\infty} (n^{n^\alpha}-1)$ ? – mr.pink Apr 05 '20 at 15:48
  • @mr.pink: It seems that you may not understand the notation $O\left(n^\alpha\log n\right)$. See Big O notation. It means that the left-hand side is bounded above by a multiple of $n^\alpha\log n$. So it's in a sense the opposite of the fact that $\mathrm e^{n^\alpha\log n}-1\ge n^\alpha\log n$, which bounds the left-hand side from below. It comes from $\mathrm e^x=1+O(x)$ as $x\to0$, which you can prove using the Taylor expansion of $\mathrm e^x$. (Thus I'm using the fact that $n^\alpha\log n\to0$ for $n\to\infty$ if $\alpha\lt-1$.) – joriki Apr 05 '20 at 16:03
  • @mr.pink: Sorry, there were some typos in the above comment; if you'd already started reading it, please reload. – joriki Apr 05 '20 at 16:06
  • @mr.pink: This is why I structured the proof in this way: First I used the general result $\mathrm e^x\gt1+x$, which is not merely asymptotic and doesn't depend on the value of $\alpha$, to establish that we only need to look at $\alpha\lt-1$; then I used $\alpha\lt-1$ to get asymptotic bounds that only hold if the exponent goes to zero. – joriki Apr 05 '20 at 16:08
  • Oh ok, so if I get it correctly the $n^{n^\alpha}-1=\mathrm e^{n^\alpha\log n}-1=O\left(n^\alpha\log n\right)\quad\text{for $n\to\infty$}$ means, that as $n\to \infty$ the original series is upper bounded by some multiple of $n^\alpha \log(n)$ and therefore we can decide on convergence by studying the convergence of $\sum_nn^\alpha\log n$ ? – mr.pink Apr 05 '20 at 16:15
  • @mr.pink: Exactly. – joriki Apr 05 '20 at 16:17