Is the series $\sum(-1)^n(2^{\frac{1}{n}} - 1)$ convergent? absolutely? Is the series $\sum(-1)^n(2^{\frac{1}{n}} - 1)$ convergent? Does it converge absolutely?
 A: The series converges. To show this we can use the alternating series test. It is easy to check that the sequence $b_n=2^{\frac{1}{n}} - 1$ is monotone decreasing. Therefore it suffices to show $\lim_{x\to\infty} (2^{\frac{1}{n}} - 1)= 0$. Notice:
$$(\lim_{x\to\infty} 2^{\frac{1}{n}} - 1)=\lim_{x\to\infty} 2^{\frac{1}{n}} - \lim_{x\to\infty} 1 =\lim_{x\to\infty} 2^{\frac{1}{n}} - 1$$
since $f(x) = 2^\frac{1}{x}$ is continuous $$\lim_{x\to\infty} 2^{\frac{1}{n}}= 2^{\lim_{x\to\infty} \frac{1}{n}} = 2^0=1$$
Therefore $$b_n \to 0$$
Hence the series converges by the alternating series test.
A: As @SamaelManasseh showed it's conditionally convergent.
But he was wrong in absolutely convergent proof! Notice he must write, $\sum(\frac{1}{n}-1)$, that he missed the parentheses. And in this case, $(\frac{1}{n}-1 < 0)$, that fails in comparison test.
But you can follow what @zwim said in comment:
$$f(t) = 2^t - 1 - \frac{t}{2}$$
$$f'(t) = 2^t\ln(2) - \frac{1}{2}$$
$$t>0 \Longrightarrow 2^t>1 \overset{\ln(2)>\frac{1}{2}}{==\Longrightarrow} 2^t\ln(2)>\frac{1}{2} \Longrightarrow f'(t)>0 \overset{f(0)=0}{==\Longrightarrow}$$
$$\overset{f(0)=0}{==\Longrightarrow} f(t)>0 \Longrightarrow 2^t - 1>\frac{t}{2} \Longrightarrow 2^{\frac{1}{n}} - 1>\frac{1}{2n}$$
That's a positive divergent series which shows your series isn't absolutely divergent by comparison test.
