Have I proved that the series diverges by the limit test?

Question

I have to find out whether the following series diverges. $$ \sum_{k=1}^{\infty}\frac{(-1)^k}{k^{1/k}} $$ I thought that it would be correct if I just write something like this: $$ \lim_{k\rightarrow\infty}\frac{(-1)^k}{k^{1/k}}=\lim_{k\rightarrow\infty}(-1)^k=A \\ \left\{ \begin{aligned} &A = \lim_{k_1\rightarrow\infty}(-1)^{2k_1}=1 \\ &A = \lim_{k_2\rightarrow\infty}(-1)^{2k_2-1}=-1 \end{aligned}\Rightarrow\nexists A\Rightarrow \text{the initial series diverges.} \right. $$ Am I right here?

Answer 1

I'd say you have exactly the right idea! The Divergence Test says that if the summand approaches anything other than $0$ (or does not exist) then the series diverges. You were able to see that the limit tends to something that flips between $-1$ and $1$ (which is not $0$) and therefore diverges by the Divergence Test.

These cases can be tricky to write up, so while your write-up may be acceptable to some, it may not be to others. For example, to say $\displaystyle\lim_{k \to \infty} \dfrac{(-1)^k}{k^{1/k}}= \lim_{k \to \infty} (-1)^k$ you either need to write more, or have taken the limit without really having taken the limit. [After all, how else did the $k^{1/k}$ term vanish?!] So a write-up like below would be cleaner:

For a limit to exist, it must exist 'any way you go to infinity'.

$$ \lim_{k \to \infty} \dfrac{(-1)^{2k}}{(2k)^{1/(2k)}}= \lim_{k \to \infty} \dfrac{1}{(2k)^{1/(2k)}}= \dfrac{1}{1}=1 $$ $$ \lim_{k \to \infty} \dfrac{(-1)^{2k+1}}{(2k+1)^{1/(2k+1)}}=\lim_{k \to \infty} \dfrac{-1}{(2k+1)^{1/(2k+1)}}= -1 $$ Therefore, the limit $\displaystyle\lim_{k \to \infty} \dfrac{(-1)^k}{k^{1/k}}$ does not exist so that $\displaystyle\sum_{n=1}^\infty \dfrac{(-1)^k}{k^{1/k}}$ diverges by the Divergence Test.

Answer 2

Yes, that's right, since for a series to converge, it is necessary that its terms tend to nothing.

But as you've rightly calculated, the terms of your series above do not go to zero.