# Determine Series' Convergence or Divergence

1. May 7, 2012

### knowLittle

1. The problem statement, all variables and given/known data
$\sum _{n=1}ne^{-n}$

2. Relevant equations
Ratio Test
Integral Test

3. The attempt at a solution
I know that by the ratio test, it converges absolutely. But, I am unable to determine its convergence through the integral test . Could someone help? I thought that the integral test would be best, since e is used.

The ratio test gives me 1/e approx 0.367879 <1; therefore, it converges absolutely.

Now, the integral test is my concern. I believe I have done all correct steps, but the result shows that it diverges.
$\lim _{b\rightarrow \infty }\int _{1}^{b}\dfrac {n} {e^{n}}dn$

Integrating by parts with: u=n; du=dn; dv=(e^-n) dn; v=-e^-n
$\left[ -e^{-n}\left( n\right) +-e^{-n}\right] _{1}^{\infty }$
The final result is:
$-\dfrac {\infty } {e^{\infty} }-\dfrac {1} {e^{\infty }}+\dfrac {2} {e}$ and it diverges. What am I doing wrong?
Thank you.

2. May 7, 2012

### scurty

What is $\displaystyle \lim_{n \rightarrow \infty} \frac{n}{e^n}$? You will need to use l'Hospital's rule!

3. May 7, 2012

### knowLittle

Wow. Thank you.

4. May 7, 2012

### knowLittle

My integral after L'Hospital doesn't agree with answers online. Could you tell me what's wrong?
$\lim _{b\rightarrow \infty }\int _{1}^{b}\dfrac {n} {e^{n}}dn$

$\int _{1}^{b}\dfrac {1} {en}dn=\left[ -e^{-n}\right] _{1}^{b}=-\dfrac {1} {e^{\infty }}+\dfrac {1} {e}$
, the first term to the left is e^n

5. May 7, 2012

### sharks

You need to find the limit of $\frac{1}{e^n}$ and not its integral.

6. May 7, 2012

### knowLittle

The limit of (1/ e^n) as n approaches infinity is 0. The test for divergence is inconclusive.

I'm trying to find the determination of convergence or divergence through the integral test.

Last edited: May 7, 2012
7. May 7, 2012

### sharks

OK, i've worked it out, and the limit should be: $\frac{2}{e}$

Last edited: May 7, 2012
8. May 7, 2012

### knowLittle

Yeah, that's what it's shown online.
I get 1/e. My procedure is above. Could you help?

9. May 7, 2012

### sharks

This step is correct:
$$\left[ -e^{-n}\left( n\right) -e^{-n}\right] _{1}^{\infty }$$
Let b=∞, then,
$$\left[ -e^{-n}\left( n\right) -e^{-n}\right] _{1}^{b}$$
This gives:
$$\left[ \left( \frac{-b}{e^b} -e^{-b}\right) - \left( \frac{-1}{e} -e^{-1}\right) \right]$$
Now, what you have to find is the limit as b→∞
$$\lim_{b→∞} \left[ \left( \frac{-b}{e^b} -e^{-b}\right) - \left( \frac{-1}{e} -e^{-1}\right) \right]$$
But the limit of the following requires using L'Hopital's rule,
$$\lim_{b→∞} \frac{-b}{e^b} = \lim_{b→∞} \frac{-1}{e^b}$$
You can then work out the rest:
$$\lim_{b→∞} \left[ \left( \frac{-1}{e^b} - \frac{1}{e^b}\right) - \left( \frac{-1}{e} - \frac{1}{e}\right) \right]$$

10. May 7, 2012

### knowLittle

Thanks, it's what I suspected. This L'Hospital rule keeps coming and coming.

11. May 7, 2012

### scurty

You had the answer in your first post! The first term goes to zero by l'Hospital's, the second goes to zero, and the final term is just a constant which was $\frac{2}{e}$.

12. May 7, 2012

### knowLittle

I know, I didn't know I could use L'Hospital in the middle of an integration by parts.
Thanks.