Why is $\sum_{n=1}^\infty {1\over{n^{1+ {1\over\ln \ln n}}}}$ divergent?

Question

I am just starting to learn Calculus. If anyone could help me that would be very useful. Thanks ahead

From here: how to prove $\sum {\frac{1}{n^{1+1/n}}}$ is divergent

I don't really get how to use induction from $\dfrac{1}{n ^ {1+ \frac{1}{n}}} \lt \dfrac{1}{2n}$

And another one is why $\sum_{n=1}^\infty {1\over {n ^ {1+ {1\over n}}}}$ is divergence and $\sum_{n=1}^\infty {1\over{n^{1+ {1\over \ln n} }}}$ Is not?

I mean $\sum_{n=1}^\infty {1\over{n^{1+ {1\over \ln (\ln n) } }}}$ (Sorry)

Answer 1

After your edit: we will show that the series with general term $\frac{1}{n^{1+\frac{1}{\ln \ln n}}}$ (for $n\geq 3$) is convergent. To do so, we will compare it to the (convergent) series $\sum_{n=2}^\infty \frac{1}{n(\ln n)^2}$, which is a convergent Bertrand series.

Rewrite $$ n^{1+\frac{1}{\ln\ln n}} = n\cdot n^{\frac{1}{\ln\ln n}}= n\cdot e^{\frac{\ln n}{\ln\ln n}} $$ and $$n(\ln n)^2 = n\cdot e^{2\ln \ln n}$$

We have asymptotically that $\ln \ln n = o\left(\frac{\ln n}{\ln\ln n}\right)$, i.e. for $n$ big enough $\frac{\ln n}{\ln\ln n} > 2\ln \ln n$. This implies $$\frac{1}{n^{1+\frac{1}{\ln\ln n}}} < \frac{1}{n(\ln n)^2}$$ for $n$ big enough, and by comparison that the series $\sum_{n=3}^\infty \frac{1}{n^{1+\frac{1}{\ln \ln n}}}$ converges.

Answer 2

HINT:

Note that we have

$$n^{1/\log(n)}=e$$

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Note that in the original post, the series of interest was $\sum_{n=2}^{\infty}\frac{1}{n^{1+\frac{1}{\log(n)}}}$