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.