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.


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 $\sum\frac1n$.

  • $\begingroup$ 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. $\endgroup$ – Zein Nov 7 '14 at 17:38
  • $\begingroup$ Do you know the Taylor series? $\endgroup$ – user63181 Nov 7 '14 at 17:58
  • $\begingroup$ 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) $\endgroup$ – Zein Nov 7 '14 at 18:00
  • $\begingroup$ 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. $\endgroup$ – user63181 Nov 7 '14 at 18:07
  • $\begingroup$ Thanks, but how do did you get that limit = ln2? $\endgroup$ – Zein Nov 7 '14 at 18:11

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.

  • 1
    $\begingroup$ $2^{\frac 1n} \to 1$ when $n \to \infty$ so the then the terms aproach $0$ $\endgroup$ – Ahlfkushevich Feb 22 at 20:46

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