# Thread: Some problems with series

1. Originally Posted by Hweengee
How do I determine the convergence/divergence of the following:

$
\sum _{n=1}^{\infty }\left({\sqrt[n] 3}-1 \right)
$

I can't use limit comparison test with ${\sqrt[n] 3}$ because it gives me 0. Root and ratio tests don't seem appropriate and I'm struggling to find a divergent series which is smaller than this to compare it with. Would it be possible to compare it with $\sum _{n=1}^{\infty }\left(\ln\left(n \right)-1 \right)$?

As for

$\sum _{n=3}^{\infty }\frac{1}{\left( lnln\left(n \right) \right)^{\ln \left( n \right) }}$

I'm more or less completely stuck.

Thanks for the help in advance.
I see that your second series was never actually solved. Well I would suggest Cauchy's Condensation test. Since the sequence in the series is positive and decreasing it only converges iff $\sum_{n=3}^{\infty}\frac{2^n}{\ln(\ln(2^n))^{\ln(2 ^n)}}=\sum_{n=3}^{\infty}\frac{2^n}{\ln(\ln(2)n)^{ \ln(2)n}}\sim\sum_{n=3}^{\infty}\frac{2^n}{\ln(n)^ n}$. Now it is pretty clear this converges.

2. would you please explain furthur that why it converge. as a glance i see that it is divergant.

3. sorry with te RATIO TEST i found it convergant

4. Originally Posted by Mathstud28
I see that your second series was never actually solved. Well I would suggest Cauchy's Condensation test. Since the sequence in the series is positive and decreasing it only converges iff $\sum_{n=3}^{\infty}\frac{2^n}{\ln(\ln(2^n))^{\ln(2 ^n)}}=\sum_{n=3}^{\infty}\frac{2^n}{\ln(\ln(2)n)^{ \ln(2)n}}\sim\sum_{n=3}^{\infty}\frac{2^n}{\ln(n)^ n}$. Now it is pretty clear this converges.
It is possible to compare with a simple series: we have $(\ln\ln n)^{\ln n}=e^{\ln n\ln\ln\ln n}=n^{\ln\ln\ln n}\geq n^2$ as soon as $n\geq e^{e^{e^2}}\simeq 5.8\cdot 10^{702}$ (!!), hence $\frac{1}{(\ln\ln n)^{\ln n}}\leq \frac{1}{n^2}$ for these pretty large $n$. Hence it converges.

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