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Does the series $\sum\limits_{n=1}^\infty (2^{1/n} - 1)=(2^1 - 1) + (2^{\frac{1}{2}} - 1)+ ... +(2^{\frac{1}{n}}-1)+...$ converge?

I feel like it diverges the same way as $\sum \frac{1}{n}$ diverges: very slow, on a logarithmic scale.

Did
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Saketh Malyala
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    Good intuition! Why do you think $2^{1/n}-1$ and $1/n$ ought to behave similarly? If you can formalise this idea, you're very close to a proof – πr8 Dec 30 '16 at 18:52

3 Answers3

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Your intuition is correct!

Using the fact that $e^x-1 \ge x$ for all real $x$, we have $2^{\tfrac{1}{n}}-1 = e^{\tfrac{1}{n}\ln 2}-1 \ge \dfrac{1}{n}\ln 2$ for all $n$.

Then, since $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{n}\ln 2$ diverges, $\displaystyle\sum_{n = 1}^{\infty}2^{\tfrac{1}{n}}-1$ diverges by direct comparison.

JimmyK4542
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Your series and the harmonic series are asymptotically equivalent, as $$ \lim_{n\to\infty}\frac{2^{1/n}-1}{1/n}=\lim_{x\to0}\frac{2^x-1}{x}=\ln2 $$

egreg
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You can also go about this problem by using the Comparison Test:

Let $\sum$ $a_n$, $\sum$ $b_n$ be 2 positive sequences where $a_n$ $\le$ $b_n$. $$\\$$ $\sum\limits_{n=1}^\infty (2^{1/n} - 1)$ $\ge$ $\sum\limits_{n=1}^\infty (-1)$

Since every infinite sum of a non-zero constant diverges, $\sum\limits_{n=1}^\infty (-1)$ diverges. Thus $\sum\limits_{n=1}^\infty (2^{1/n} - 1)$ diverges.

BBot
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  • You're missing the rest of the comparison test. But you've also misapplied it. The rightmost sum diverges to $-\infty$, so you've shown that the OP's sum either converges or diverges (either to $+\infty$ or it will cycle). – Harry Jul 11 '17 at 05:42
  • This can only be used when $b_n$ is positive – Guanyuming He Apr 08 '21 at 13:46