# I want to see if I'm right...

• Jan 15th 2008, 05:21 AM
Undefdisfigure
I want to see if I'm right...
I am given the series (e^n)/(3^(n-1))...Don't forget the Riemann sum symbol in front of this. It appears to be a geometric series. This would mean its ar^n
and a = 1/3 and r = e. I then do a bit of manipulation and get 1/3-1/3(e^n) and then Sn = (1-e^n)/(3-3e)

since r>1, the sequence {r^n} is divergent. Lim n--> infinity Sn does not exist. The geometric series is divergent.
• Jan 15th 2008, 05:39 AM
colby2152
Quote:

Originally Posted by Undefdisfigure
I am given the series (e^n)/(3^(n-1))...Don't forget the Riemann sum symbol in front of this. It appears to be a geometric series. This would mean its ar^n
and a = 1/3 and r = e. I then do a bit of manipulation and get 1/3-1/3(e^n) and then Sn = (1-e^n)/(3-3e)

since r>1, the sequence {r^n} is divergent. Lim n--> infinity Sn does not exist. The geometric series is divergent.

$\sum_{n=0}^{\infty}3(\frac{e}{3})^n = \frac{3}{1-\frac{e}{3}} \Rightarrow \frac{9}{3-e}$

It is a geometric series - how is it divergent?
• Jan 15th 2008, 06:37 AM
topsquark
Quote:

Originally Posted by Undefdisfigure
I am given the series (e^n)/(3^(n-1))...Don't forget the Riemann sum symbol in front of this. It appears to be a geometric series. This would mean its ar^n
and a = 1/3 and r = e. I then do a bit of manipulation and get 1/3-1/3(e^n) and then Sn = (1-e^n)/(3-3e)

since r>1, the sequence {r^n} is divergent. Lim n--> infinity Sn does not exist. The geometric series is divergent.

Your mistake appears to be in taking the limit of your sum formula:
$\sum_{i = 0}^n \frac{e^i}{3^{i - 1}}$

Notice that your r is not e, but e/3 < 1. Thus when we take
$\lim_{n \to \infty} \sum_{i = 0}^n \frac{e^i}{3^{i - 1}} = \lim_{n \to \infty} \sum_{n = 0}^n 3 \left ( \frac{e}{3} \right )^i$

$= \lim_{n \to \infty} \frac{3 \left (1 - \left [ \frac{e}{3} \right ] ^{n + 1} \right )}{1 - \frac{e}{3}}$

$= \frac{3}{1 - \frac{e}{3}}$
(the e/3 term in the numerator vanishes because e/3 < 1)

$= \frac{9}{3 - e}$

-Dan
• Jan 15th 2008, 04:02 PM
Undefdisfigure
I have another question. I have to find a telescoping sum of ln n/(n+1). I do not see how to do a partial fraction decomposition on this. Do I have to take the derivative of this expression? I have to find if this expression is convergent and if it is, to find its sum.
• Jan 15th 2008, 04:06 PM
Jhevon
Quote:

Originally Posted by Undefdisfigure
I have another question. I have to find a telescoping sum of ln n/(n+1). I do not see how to do a partial fraction decomposition on this. Do I have to take the derivative of this expression? I have to find if this expression is convergent and if it is, to find its sum.

what partial fraction? recall, $\log_a \left( \frac xy \right) = \log_a x - \log_a y$

Aside: ask new questions in a new thread