Root Test: Convergence of Series

[azatkgz]

Homework Statement

Determine whether the series converges or diverges.

\sum_{n=2}^{\infty}\frac{1}{(\ln n)^{\ln (\ln n)}}=\sum_{n=2}^{\infty}\frac{1}{e^{\ln (\ln n)\ln (\ln n)}}

The Attempt at a Solution

for u=\ln (\ln n)

\sum_{u=\ln (\ln 3)}^{\infty}\frac{1}{e^{u^2}}

from Root Test

\lim_{u\rightarrow\infty}\sqrt<u>{\frac{1}{e^{u^2}}}=\lim_{u\rightarrow\infty}\frac{1}{e^u}=0&lt;1</u>
series converges

 

[quasar987]

That u substitution does not give back the original series. ln(ln(3))+1 does noe equal ln(ln(4)).

Try simply the comparison criterion. Hint: (logx)²/x -->0, so (logx)²<x for x large enough.

 

[azatkgz]

Try simply the comparison criterion. Hint: (logx)²/x -->0, so (logx)²<x for x large enough.

\ln^2x&lt;x

for x=lnn

\ln^2(\ln n)&lt;\ln n

e^{\ln^2(\ln n)}&lt;e^{\ln n}=n

e^{\ln^2(\ln n)}&lt;n\rightarrow \frac{1}{e^{\ln^2(\ln n)}}&gt;\frac{1}{n}

diverges

 

[quasar987]

good!