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

Solution 1:

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

Solution 2:

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

Solution 3:

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

Solution 4:

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.

Solution 5:

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} $.