Root Test: Convergence of Series

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
azatkgz
Messages
182
Reaction score
0

Homework Statement


Determine whether the series converges or diverges.

[tex]\sum_{n=2}^{\infty}\frac{1}{(\ln n)^{\ln (\ln n)}}=\sum_{n=2}^{\infty}\frac{1}{e^{\ln (\ln n)\ln (\ln n)}}[/tex]





The Attempt at a Solution



for [tex]u=\ln (\ln n)[/tex]

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

from Root Test

[tex]\lim_{u\rightarrow\infty}\sqrt<u>{\frac{1}{e^{u^2}}}=\lim_{u\rightarrow\infty}\frac{1}{e^u}=0<1</u>[/tex]
series converges
 
Physics news on Phys.org
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.
 
quasar987 said:
Try simply the comparison criterion. Hint: (logx)²/x -->0, so (logx)²<x for x large enough.

[tex]\ln^2x<x[/tex]

for x=lnn

[tex]\ln^2(\ln n)<\ln n[/tex]

[tex]e^{\ln^2(\ln n)}<e^{\ln n}=n[/tex]

[tex]e^{\ln^2(\ln n)}<n\rightarrow \frac{1}{e^{\ln^2(\ln n)}}>\frac{1}{n}[/tex]

diverges