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

Question

I need to argue if this series converges or diverges.

I know that

$$\sum_{n=1}^{\infty} 2^{\frac{1}{n}} = \infty$$

and

$$\sum_{n=1}^{\infty} -2^{\frac{1}{n+1}} = \infty$$

But my first impression is that

$$\sum_{n=1}^{\infty} (2^{1/n} - 2^{1/(n+1)})$$

converges as the factor $(2^{1/n} - 2^{1/(n+1)})$ seems decreasing at every iteration of $n$, I know that if a serie $\sum_{n=1}^{\infty} a_n$ is convergent, then $lim_{n\to \infty} a_n = 0$, I already proved that $\lim_{n\to \infty}(2^{1/n} - 2^{1/(n+1)})=0$ but I know I can't conclude that the serie is convergent, and I was trying to use the comparison theorem to prove its convergence but I'm stuck at what series should I compare.

Answer 1

N-th partial sum of the series is $\frac 1 2 -\frac 1 {2^{N+1}}$ which tends to $\frac 1 2 $.

Answer 2

Note that

$$2^{1/n} =e^\frac{\log 2}n=1+\frac{\log 2}n+O\left(\frac1{n^2}\right) $$

$$2^{1/n+1} =e^\frac{\log 2}{n+1}=1+\frac{\log 2}{n+1}+O\left(\frac1{n^2}\right) $$

therefore

$$2^{1/n} - 2^{1/n+1}= \frac{\log 2}n-\frac{\log 2}{n+1}+O\left(\frac1{n^2}\right)=\frac{\log 2}{n(n+1)}+O\left(\frac1{n^2}\right)=O\left(\frac1{n^2}\right)$$

the refer to limit comparison test with $\sum \frac1{n^2}$.

Answer 3

It's a telescoping sum, and the partial sum is $2^{1/1}-2^{1/(N+1)}$.
Thus the limit gives $2-1=1$