Is $\sum_{n=1}^\infty(2^{\frac1{n}}-1)$ convergent or divergent? $$\lim_{n\to\infty}(2^{\frac1{n}}-1) = 0$$ I can't think of anything to compare it against. The integral looks too hard: $$\int_1^\infty(2^{\frac1{n}}-1)dn = ?$$ Root test seems useless as $\left(2^{\frac1{n}}\right)^{\frac1{n}}$ is probably even harder to find a limit for. Ratio test also seems useless because $2^{\frac1{n+1}}$ can't cancel out with ${2^{\frac1{n}}}$. It seems like the best bet is comparison/limit comparison, but what can it be compared against?

  • $\begingroup$ @Alex That does not work, note the $-1$ after the $2^\frac1n$ $\endgroup$ – Justin Apr 27 '13 at 0:19
  • $\begingroup$ Wow, I seem to be getting good at asking questions. ($3$rd question with high votes) $\endgroup$ – Justin Apr 27 '13 at 0:26
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    $\begingroup$ I appreciate that your questions are formatted well, titled well, labeled correctly, and that you show some work and thought process. Please keep up the good work, and +1 $\endgroup$ – davidlowryduda Apr 27 '13 at 2:38

Elementary method: use AM $\ge$ GM!

$n-2$ copies of $1$, two copies of $\frac{1}{\sqrt{2}}$ gives

$$ \frac{n-2 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}}{n} \ge \left(\frac{1}{2}\right)^{1/n}$$

$$\frac{n-c}{n} \ge \left(\frac{1}{2}\right)^{1/n}$$

for some $c \gt 0$ ($ c= 2 - \sqrt{2}$)

$$ 2^{1/n} \ge \frac{n}{n-c} = 1 + \frac{c}{n-c}$$

Thus $$2^{1/n} - 1 \ge \frac{c}{n-c}$$

and so the series diverges.

But more simply:

$$2^{1/n} = e^{\log 2/n} \ge 1 + \frac{\log 2}{n}$$

(using $e^x \ge 1 + x$).

  • $\begingroup$ Two great solutions, wish I could give two upvotes. $\endgroup$ – vadim123 Apr 27 '13 at 0:33
  • $\begingroup$ @vadim123: You are very kind. Thank you! $\endgroup$ – Aryabhata Apr 27 '13 at 4:02
  • $\begingroup$ But why do the $n-2$ copies of $1$ and two copies of $1/2$ give $\sqrt[n]{1/2}$ and not $\sqrt[n]{1/4}$ when multiplied out $\endgroup$ – Spine Feast Jan 30 '15 at 17:41
  • $\begingroup$ @DepeHb: Uh. You are right! Thanks for pointing that out. That is easily fixed, fortunately (take $\frac{1}{\sqrt{2}}$ instead)... $\endgroup$ – Aryabhata Jan 30 '15 at 18:31

See: $\sum_{n=1}^\infty(2^{\frac1{n}}-1)$

There are a few methods listed there, one being writing ${2^{\frac1n}}$ as a power series. The easiest to understand is probably the limit comparison test where $b_n = \frac1n$.

Paraphrasing Bjartr:

Let $m = \frac{1}{n}$, then we have $$\lim_{m\rightarrow0}\frac{2^m - 1}{m} = \frac{0}{0}$$ So we use L'hopital's Rule $$\lim_{m\rightarrow0}2^m\log(2) = \log(2) \neq 0$$ So $\sum_{n=1}^\infty(2^{\frac1{n}}-1)$ has the same behavior as $\sum_{n=1}^{\infty}\frac{1}{n}$ which diverges. Therefore: $\sum_{n=1}^\infty(2^{\frac1{n}}-1)$ is divergent

  • $\begingroup$ Why asking when you already knew the answer? $\endgroup$ – Shuhao Cao Apr 27 '13 at 0:11
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    $\begingroup$ "Share your knowledge, Q&A-Style". $\endgroup$ – Justin Apr 27 '13 at 0:13

Try the Comparison Test, using the elementary inequality $$ 2^{1/n} -1 > {\log 2\over n} $$ for $n=1,2,\ldots$.

  • $\begingroup$ $2^{1/n} = e^{(\log 2)/n} = 1 + (\log 2)/n + $ positive terms. $\endgroup$ – GEdgar Apr 27 '13 at 0:53

No need for L'hopital: Just use the fact that by the definition of derivative, $$\lim_{x \rightarrow 0} {2^x - 1 \over x} = {d \over dx} 2^x\bigg|_{x = 0}$$ $$ = \ln 2$$ So as in bjatr's answer, this means that $$\lim_{n \rightarrow \infty} {2^{1 \over n} - 1 \over {1 \over n}} = \ln 2$$ So by the limit comparison test, the series diverges.


Yet another elementary method

$1=2-1=(2^\frac{1}{n})^n-1=(2^\frac{1}{n}-1)(2^\frac{n-1}{n}+2^\frac{n-2}{n}+\cdots+2^\frac{1}{n}+1) < (2^\frac{1}{n}-1)*(2+2+\cdots+2)$ .

Therefore, $(2^\frac{1}{n}-1) > \frac{1}{2n} $.


Since $2^x=1+x\ln 2+O(x^2)$ as $x\to 0$ then $$\sum_{n\ge 1}\left(2^{1/n}-1\right)\asymp \sum_{n\ge 1}\frac{1}{n},$$ which diverges.


You could use the fact that for a series of positive terms, $\sum_{n=1}^\infty a_n$ converges if and only if $\prod_{n=1}^\infty (1+a_n)$ converges.

Applying this result to the given problem: The given series converges if and only if the infinite product $\prod_{n=1}^\infty 2^{\frac1n}$

For this infinite product, the partial products are of the form $2^{1+\frac12+\frac13+\cdots+\frac1n}$, which is divergent since the exponent is a partial sum of the harmonic series, and hence going to $\infty$.


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