# Series: converges or diverges help

1. Apr 6, 2010

### Slimsta

1. The problem statement, all variables and given/known data
Determine whether the series [PLAIN]http://img265.imageshack.us/img265/69/44751948.jpg [Broken] converges or diverges and find its sum S if it does converge.

2. Relevant equations

3. 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?

Last edited by a moderator: May 4, 2017
2. Apr 6, 2010

### rs1n

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 $\ln\left(\frac{2n+1}{3n-1}\right)$

Last edited by a moderator: May 4, 2017
3. Apr 6, 2010

### lanedance

worth checking what the nth term does for large n

4. Apr 6, 2010

### Slimsta

$\frac{2n+1}{3n-1}$ 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'?

5. Apr 6, 2010

### Char. Limit

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

6. Apr 7, 2010

### Slimsta

ln(1) = 0

so the answer is Diverges, S=0. ?

7. Apr 7, 2010

### Staff: Mentor

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

8. Apr 7, 2010

### 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.

Last edited: Apr 7, 2010
9. Apr 7, 2010

### rs1n

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.