Series: converges or diverges help

[Slimsta]

Homework Statement

Determine whether the series [PLAIN]http://img265.imageshack.us/img265/69/44751948.jpg converges or diverges and find its sum S if it does converge.

Homework Equations

The Attempt at a Solution

my textbook has lots of weird examples but none of them help me understand how to solve this one..
if i plug in 1 then 2 then 3... i get
ln(1.5) + ln(1) + ln (7/8) +...+ln((2n+1)/(3n-1))

so a = ln(1.5)
r = ?

then i use the formula: a / (a-r)
right? now how do i find r?

 

[rs1n]

Homework Statement

Determine whether the series [PLAIN]http://img265.imageshack.us/img265/69/44751948.jpg converges or diverges and find its sum S if it does converge.

Homework Equations

The Attempt at a Solution

my textbook has lots of weird examples but none of them help me understand how to solve this one..
if i plug in 1 then 2 then 3... i get
ln(1.5) + ln(1) + ln (7/8) +...+ln((2n+1)/(3n-1))

so a = ln(1.5)
r = ?

then i use the formula: a / (a-r)
right? now how do i find r?

What can you say about \frac{2n+1}{3n-1} as n\to\infty? And then about \ln\left(\frac{2n+1}{3n-1}\right)? Now considering adding a bunch of terms that eventually behave like \ln\left(\frac{2n+1}{3n-1}\right)

 

[lanedance]

worth checking what the nth term does for large n

 

[Slimsta]

What can you say about \frac{2n+1}{3n-1} as n\to\infty? And then about \ln\left(\frac{2n+1}{3n-1}\right)? Now considering adding a bunch of terms that eventually behave like \ln\left(\frac{2n+1}{3n-1}\right)

\frac{2n+1}{3n-1} as n\to\infty will be 0
and for \ln\left(\frac{2n+1}{3n-1}\right) it will be ln(0) which is impossible..

but if plugging numbers like 10, 100, 1000... into \ln\left(\frac{2n+1}{3n-1}\right) the number becomes smaller and smaller all the way to -\infty
so does it mean that the answer is 'Diverges'?

 

[Char. Limit]

\frac{2n+1}{3n-1} as n\to\infty will be 0
and for \ln\left(\frac{2n+1}{3n-1}\right) it will be ln(0) which is impossible..

but if plugging numbers like 10, 100, 1000... into \ln\left(\frac{2n+1}{3n-1}\right) the number becomes smaller and smaller all the way to -\infty
so does it mean that the answer is 'Diverges'?

As n goes to infinity, \frac{2n+1}{3n-1} does not go to zero.

 

[Slimsta]

As n goes to infinity, \frac{2n+1}{3n-1} does not go to zero.

oh my bad.. its 1
ln(1) = 0

so the answer is Diverges, S=0. ?

 

[Mark44]

oh my bad.. its 1
ln(1) = 0

so the answer is Diverges, S=0. ?

Wow! Two mistakes in only three lines!

(2n + 1)/(3n - 1) doesn't approach 1, so ln( (2n + 1)/(3n - 1)) doesn't approach 0.

Furthermore, in a series \sum a_n, if lim a_n = 0, you can't conclude much of anything. Consider \sum 1/n and \sum 1/n^2. In both series a_n --> 0, but the first series diverges and the second series converges.

 

[Ulagatin]

Sorry. This is incorrect.

I can not speak too much about series convergence/divergence, but I can talk about a sequence:

What happens to \frac{2n+1}{3n-1} as n ---> infinity?

You can rig this so that the 1's cancel and you are then left with \frac{2n}{3n} which obviously then cancels down to \frac{2}{3} (i.e. the limit of the sequence).

Now as you are actually dealing with a natural log, the limit of the sequence in actual fact is ln(\frac{2}{3})

If you wished to prove this you could do the following for all n > N: |ln (\frac{2n+1}{3n-1}) - ln(\frac{2}{3})|<\epsilon Also, note that this sum is not a geometric progression: you find r as such: r = \frac{t_{n}}{t_{n-1}} where t_{n} is the nth term of the series. Take note of the second term of the sequence, it equals zero, and division by zero is undefined. Also, r is the common ratio. So the series is, by definition, not geometric in nature.

I am not absolutely sure of this, but I believe the series in fact diverges to negative infinity, because of the simplification of the natural log as per above. You have an infinite number of ln(2/3) to multiply out which is equal (in a very loose sense) to negative infinity. I may be wrong however.

 

[rs1n]

oh my bad.. its 1
ln(1) = 0

so the answer is Diverges, S=0. ?

The series diverges, but not due the reasons you pointed out (which were not only incorrect).

Think about what happens if the n-th term in your sum does not tend to 0. Why does the following series diverge?

\sum_{n=1}^\infty k = k + k + k + k + k + \dotsm

where k\not=0. Think about how this relates to the series in the original problem.