1
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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 $\sum\frac1n$.

$\endgroup$
  • $\begingroup$ 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
-2
$\begingroup$

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.

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.