Question

Hey guys, I have a series I need to determine if it converges or diverges.

sigma (from n->infinity) of (ln(n))/(1.01)^n.

I set b_n = 1/1.01^n, conclude its convergence, by geometric series. Then I do Limit comparisin test of lim a_n/b_n and get infinity as the answer. Now the rule states that if limit gives greater than 0, it should either converge or diverge. Does this rule include infinity as the answer, and if so, is my final answer that both a_n and b_n converge.

Answer 1

\[\large\sum_{n=1}^{\infty}\frac{\ln(n)}{(1.01)^n}\]
Is this the series?

Answer 2

yes that is the series and I think I figured it out. I used Ratio test instead of limit comparison. lim a_n+1/a_n gave a limit of 1/1.01, and according to the rules it is less than 1 so converges.

Answer 3

Yeah, it converges via root and ratio. The rest are inconclusive.

Answer 4

406.224 is what i got as for where it converges.

Answer 5

you mean the limit or the actual sum S? Also for future reference, in limit comparison test, if I get infinity in the limit, does it mean it is inconclusive.

Answer 6

Yeah, the sum.
And yeah infinity is an inconclusive answer for limit comparison.

Answer 7

Thank you my friend

Answer 8

Actually, now that I remember entirely, infinity is an answer for the limit comparision test... it implies both are divergent.
You cant use limit comparision on this though. The Lemma for it is:
0 < lim n->infty (a_n/b_n) < infinity

and:
\[\lim_{n\rightarrow\infty}\frac{\ln n}{(1.01)^n}=0\]

Answer 9

I see thanks for that.