Question

Need some help doing limits of series. Will post question.

Answer 1

\[\sum_{n=1}^\infty \ln(2(n+1)) - \ln(2n)\]
What is the first step to solving this? Thank you in advance :)

Answer 2

well,,you're welcome first of all! :P

now,we have
ln(2(n +1)) - ln(2n)

you much be knowing log(ab) = log a + log b

we just apply that here to get
ln 2 + ln (n+1) - ln2 - ln(n)
= ln(n+1) - ln(n)

just keep on putting values of n 1 by 1..like this

ln(2) - ln(1) + ln(3) - ln(2) + (ln4) - ln3 .......

you see?

Answer 3

ah ok so I used log property to get it into the form ln(n+1)-ln(n). Is it like taking a limit now? Sorry if it's a stupid question... it's pretty late xD.

Answer 4

you have summation right?
summation n from 1 to infinity ..

just keep on adding putting values of n one by one..

Answer 5

How do I know if it converges to a number though? What if it just goes on forever?

Answer 6

doesn't this one go onto infinity?

Answer 7

its 0,ofcorse,,every number gets cancelled out..

Answer 8

oh ok so everything cancels? (what about the -ln(1)?) That doesn't cancel, does it?

Answer 9

ln1 = 0

Answer 10

ohhhhhh duh... ok that makes sense.

Answer 11

Thank you :D

Answer 12

glad to help :)