Question

For which \(a\) does the following series converge?

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

Answer 1

For which \(a\) does the following series converge?

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

Answer 2

i think for "a" from (0,+inf)

Answer 3

does not converge for a=1

Answer 4

I can tell you that it converges for a>= 2 but I don't know the reason yet

Answer 5

We might have to use comparison test

Answer 6

integral test

Answer 7

have you figured it out yet :)

Answer 8

foods ready..be back

Answer 9

a>1

Answer 10

\[a_n=\left(\frac{1}{n*\text{Log}[n]^2}\right)\]
\[b_n=\frac{1}{n^2}\]

\[\frac{a_n}{b_n}\] as n goes to infinity
\[\frac{\left(\frac{1}{n*\text{Log}[n]^2}\right)}{\frac{1}{n^2}}\]
\[n\rightarrow \infty \]\[\frac{n}{\text{Log}[n]^2}\] =0

According to rules two of limit comparison test , it converge
http://archives.math.utk.edu/visual.calculus/6/series.9/index.html

Answer 11

is should converge for a=2 since it converges for all a>1

Answer 12

a>1, bingo. Thanks!

Answer 13

@imranmeah91: that limit above is not zero

Answer 14

what do you mean?

Answer 15

\[\lim_{n\to\infty}\frac{n}{\text{Log}[n]^2}\neq0\]

Answer 16

nvm, got it