Question

show the series with term ln r/r^2 converges

Answer 1

i.e. \[\ln r \over r^2\]

Answer 2

i figure if this term is \[u _{n}\] then if i can find \[a_{n}\] that is < than \[u _{n}\] that also converges i will be able to show it; however 1/r^2 doesn't get me there and I can't show it for 1/r^3, hence stumped

Answer 3

i've seen a few questions with ln r in numerator, so i'm guessing this is a common thing

Answer 4

\[\lim_{n \rightarrow \infty} \frac{\ln}{n ^{2}}\]\[\lim_{n \rightarrow \infty} \frac{1/n}{2n}\]
\[\lim_{n \rightarrow \infty} \frac{-1/n ^{2}}{2}\]
=0

Answer 5

what is ln/n^2

Answer 6

I made a mistake, it is lnn/n^2

Answer 7

i don't follow your logic there, are you comparing with the term 1/2n^2

Answer 8

do you know l`hopital`s rule

Answer 9

no

Answer 10

its not mentioned in my further math book

Answer 11

i can only assume its possible without it

Answer 12

ok it is says that if there is \[\frac{\infty}{\infty}\]

you can derivative proportion and denominator

Answer 13

oh,thats interesting

Answer 14

i'll google it

Answer 15

thanks for your help, odd that the book would ask the qu and no mention of the rule

Answer 16

i think the answer is to compare with r^0.5

Answer 17

i.e. as r^0.5/r^2 converges ( this is the same as 1/r^1.5)
and ln r < r^0.5 for all r>0