sum of the series $(1+n)^{1/5} - n^{1/5}$ as $n$ tends to infinity

Question

The sum of the series ${(1+n)}^{1/5} - n^{1/5}$ when $n$ goes from 0 to infinity is (A) less than -1 (B) equal to -1 (C) greater than 1 but less than 2 (D) none of the above

When I expand the summation, all the terms of the series cancel each other and what we are finally left with is $(1+n)^{1/5} - 1^{1/5}$. Here I have assumed $1 = 1^{1/5}$ and the series is $S = (1+n)^{1/5} - 1 $ now $n>0$ so $1+n > 1$ that implies $(1+n)^{1/5} > 1 $ so, $(1+n)^{1/5} - 1 > 0 $ If n goes to infinity then this sum is greater than 0 but we cannot be sure that it is less than 2. so, I think (D) is the correct option. Am I correct ? please help me!
Thanks.

Answer 1

You got a correct answer.

By the telescoping summation: $$\sum_{n=1}^{+\infty}(\sqrt[5]{n+1}-\sqrt[5]{n})=\lim_{n\rightarrow+\infty}(\sqrt[5]{n+1}-1)=+\infty.$$

Answer 2

As an alternative to telescoping, we have that

$${(1+n)}^{1/5} - n^{1/5}=n^{1/5}\left[\left(1+\frac1n\right)^\frac15-1\right]=n^{1/5}\left(\frac1{5n}+O\left(\frac1{n^2}\right)\right)=\frac1{5n^\frac45}+O\left(\frac1{n^\frac95}\right)$$