Question

I have this series, and I was wondering why, when using the direct comparison test, whould i compare it to 1/n^(3/2) instead of 1/n^2. From what I have read online it seems that when using the direct comparision test, we are looking for the highest power of n in the series. So from the series I have, it seems that I should use 1/n^2 as the comaprison series. but my book uses 1/n^(3/2).

Answer 1

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

Answer 2

does it converge?

Answer 3

by using 1/n^2 and 1/n^(3/2) the series does converge

Answer 4

but why does my book use 1/n^(3/2)

Answer 5

well, direct comparison test states that in order for the series to converge, \(\large a_n \le b_n\) and \(\large \sum b_n\) is converges. In this case, your \(\large b_n = \frac{1}{n^2}\) and we know that \(\large \frac{1}{n^2} \) is convergent by the power series. \( 2 > 1 \therefore C\). So we can conclude, AND prove that this series converges.

Answer 6

Oh wait I think i messed up..........

Answer 7

So I guess they used \(\large \frac{1}{n^{3/2}} = a_n\) instead of \(\large \frac{1}{n^2}\) to show that \(\large a_n \le b_n\) ?

Answer 8

yup your right, by using b{n}=1/n^(3/2) you can say that is greater than a{n}=ln(n)/n^2

Answer 9

:) cool!

Answer 10

cause is used 1/n^2 as b{n}, it does converge,, but ln(n)/n is greater than 1/n^2 so that wouldnt be helpful

Answer 11

Yeah, you got it.

Answer 12

thank you

Answer 13

No problemo :) and thank YOU!

Answer 14

did show that \[\frac{ln(n)}{n^2}<=\frac{1}{n^{2/3}}\]

Answer 15

did you show*