Question

does [(-1)^2n ln(n)]/n converge, converge absolutely, or diverge?

Answer 1

terms go to zero, and it alternates

Answer 2

i assume again you mean the sum
if it alternates, and the terms go to zero, that is enough for convergence

Answer 3

sure as hell won't converge absolutely though, because without the \((-1)^n\) you get
\[\sum\frac{\ln(n)}{n}\]

Answer 4

here the degree of the denominator is only 1, so it is larger than
\[\sum\frac{1}{n}\]which does not converge

Answer 5

\((-1)^{2n}=1\) for all \(n\), unless that's a mistake in your series.

If it's not, you have \(\dfrac{\ln n}{n}\), where \(\ln n<n\) for \(n>1\), and thus a monotonically decreasing sequence of positive terms. Apply the integral test.

\[\sum_{n=1}^\infty \frac{\ln n}{n}~\text{converges/diverges if}~\int_1^\infty \frac{\ln x}{x}~dx~\text{converges/diverges, respectively.}\]

Answer 6

oh lord i am blind

Answer 7

on the other hand, in my defence, who writes \((-1)^{2n}\)??

Answer 8

Beats me... which is why I think it might be a mistake, or a test.

Answer 9

i wouldn’t use the integral test though, i would use the eyeball test

Answer 10

So how would you solve it?

Answer 11

If you're using the integral test, compute the integral. I'm not familiar with the eyeball test myself...

Answer 12

\[\frac{\ln(n)}{n}>\frac{1}{n}\]

Answer 13

^^^There you go.

By the way, even if it is meant to say (-1)^n, WA says it converges...
http://www.wolframalpha.com/input/?i=Sum%5B%28-1%29%5En*Log%5Bn%5D%2Fn%2C%7Bn%2C1%2CInfinity%7D%5D

Answer 14

eyeball test: denominator is a polynomial of degree only 1, so it does not converge