Convergent or Divergent? $\sum_{n=1}^\infty\bigl(2^{\frac1{n}}-1\bigr)$ 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?
 A: 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

A: Try the Comparison Test, using the elementary inequality
$$
2^{1/n} -1 > {\log 2\over n}
$$
for $n=1,2,\ldots$.
A: 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.
A: 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} $.
A: 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 $e^x \ge 1 + x$).
A: 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.
A: 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 $\infty$.
