Series question with logarithms

Question

I want to know how to check the divergence of following sum: $\sum_{k=0}^\infty \frac{1}{\sqrt[n]{\log n}}$

I tried to use this result: $ \lim_{n \rightarrow \infty} \frac{1}{\sqrt[n]{\log n}}=1 \neq 0 $ but I don't know why this result is true.

Best regards.

Answer 1

First, I assume that the index of summation is $n$ (not $k$), otherwise the sum diverges trivially. Otherwise, note that (for $n>1$) we have $n^n>n>\log n$. Hence $n>\sqrt[n]{n}>\sqrt[n]{\log n}$. We take reciprocals to get $$\frac{1}{n}<\frac{1}{\sqrt[n]{\log n}}$$

Now you can use the comparison test to complete the problem.

Answer 2

Hint: Use the Contraposition of result

if a series $\sum_{n}a_n$ converges then $\lim_{n\to \infty} a_n =0$.

See my answer.

Answer 3

When $n$ is large, $$\frac{1}{\sqrt[n]{\log n}}\simeq 1-\frac{\log (\log (n))}{n}$$ So, as you already noticed $$\lim_{n \rightarrow \infty} \frac{1}{\sqrt[n]{\log n}}=1$$ and then the sum $$\sum_{n=0}^\infty \frac{1}{\sqrt[n]{\log n}}$$ diverges.

As said in other answers, the summation should probably start at $n=2$ and not $n=0$ ($n=0$ does not make a problem but $n=1$ makes a serious one).

Using some approximations, it could be easy to show that, for $p \gt 1000$, $$\sum_{n=2}^{n=p} \frac{1}{\sqrt[n]{\log n}}\simeq p$$