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I've been working on this problem for a while, but I can't really get it. I get it, but I don't actually get it.

The question is to find whether or not this series converges: $\displaystyle\sum_{n=1}^\infty(2 ^{1/n} - 1) $

I am almost certain that it diverges, but the way I did it proved convergence, and looking into it online didn't give results I could understand, but generally agreed that it was divergent.

Let $f(x) = 2^{1/x} - 1$

The way I did it was using a comparison test of a p series $1/x^{1.1}$. Since I know $1/x^{1.1}$ is greater than $f(x)$ (this is apparently wrong, as a source online claims that $2^{1/x} - 1 \ge 1/n$) and because $p>1, 1/x^{1.1}$ converges, which would therefore, by the limit comparison test, mean that $f(x)$ must also converge since it's less than $1/x^{1.1}$.

I am at a disagreement with $1/x$ being less than or equal to $f(x)$ because when I graphed it, it was greater. $1/x^{1.1}$ was also greater when graphed.

If anyone can shed light on how $1/x$ is less than or equal to $f(x)$, that'd be great, since then by that comparison if $1/x\ge f(x)$, $f(x)$ would also diverge.

Hippalectryon
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Zein
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2 Answers2

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Notice that

$$2^{1/n}-1=\exp\left(\frac1n\ln2\right)-1\sim_\infty\frac{\ln2}{n}$$ so the series is divergent by comparison with the harmonic series $\sum\frac1n$.

  • Hey, you're someone who answered the same question a while ago, and I didn't really understand the notiation. I understand e^(ln2)/n - 1, but what does this mean? ∼∞ln2n

    I've never seen this notation.

     – Zein Nov 07 '14 at 17:38
  • Do you know the Taylor series? –  Nov 07 '14 at 17:58
  • I do not, don't think my class has gotten to that yet. This was one of those challenge questions he gave us. (Don't know power series either, if that's relevant) – Zein Nov 07 '14 at 18:00
  • We can answer the question without using the Taylor series: since $$\lim_{n\to\infty}\frac{2^{1/n}-1}{\frac1n}=\ln2$$ then we have for $n$ sufficiently large $$2^{1/n}-1\ge \frac{\ln2}{2}\frac1n$$ so we conclude the same result by comparison. –  Nov 07 '14 at 18:07
  • Thanks, but how do did you get that limit = ln2? – Zein Nov 07 '14 at 18:11
  • Change the variable with $x=\frac1n$ and then use the L'Hôpital's rule. –  Nov 07 '14 at 18:14
  • Thanks! I got it now, but there is one thing I still don't see. Why did you make it over 1/n when doing the limit in the first place? – Zein Nov 07 '14 at 19:09
  • To compare two sequence usually we look for the limit of their quotient and in your example we need to compare $2^{\frac1n}-1$ and $\frac1n$. –  Nov 07 '14 at 19:16
  • Thank you! Last thing, is (ln2)/2n an observational inequality, or is there an actual reason? – Zein Nov 07 '14 at 20:52
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Show the terms don't even approach zero (they approach 1) as n goes to infinity. Then I think it's just usually called the divergence test that is applied.

Derek Eby
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