Convergence of Summation n/e^n using L'Hospital's Rule | Homework Help

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In summary, the problem is to determine whether the series Ʃ n/e^n converges or diverges. The attempted solution involved using the integral test, but there was a sign error and the integral should have been -n/e^n - 1/e^n. The ratio test was then used, which resulted in a limit of 1/e, indicating that the series converges.
  • #1
EEintraining
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Homework Statement


Ʃ n/e^n converge or diverge


Homework Equations





The Attempt at a Solution



I got this down to an improper integral using the integral test but I am weak at L'Hospitals rules and I was wondering if someone could help me out

I have

[itex]\int n/e^n[/itex] from 1 to infinity

down to

Limit b to infinity ne^n - e^n |from 1 to b

this gives me infinity - infinity so time for L'Hospitals ( forgive my spelling)
I know to divide by the recripical of either one but I get stuck from there
 
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  • #2
It's probably easier to use the ratio test. If you have to use the integral test you didn't get the integral quite right. There's a sign problem (which isn't terribly important) and don't you mean e^(-n) in the integral (which is terribly important)?
 
  • #3
ok i will try the ratio test for this... I am studying for a test and have already worked this problem and turned it in when I got it back graded the only comments were that it was infinity-infinity and needs L'Hospitals rule. I had originally put infinity - infinity so it diverges. The original problem is correct it is n / e^n not e ^-n
 
  • #4
EEintraining said:
ok i will try the ratio test for this... I am studying for a test and have already worked this problem and turned it in when I got it back graded the only comments were that it was infinity-infinity and needs L'Hospitals rule. I had originally put infinity - infinity so it diverges. The original problem is correct it is n / e^n not e ^-n

No, I meant your integral should have been -n/e^n - 1/e^n or -ne^(-n) - e^(-n). It's not infinity-infinity.
 
  • #5
Ok I did the Ratio test could you please check my work?

[itex]\frac{n}{e^n}[/itex]

[itex]\frac{n+1}{e^(n+1)}[/itex] * [itex]\frac{e^n}{n}[/itex]

so all e's cancel except 1 giving me

[itex]\frac{n+1}{e*n}[/itex] the limit of this is ∞/∞ with L'Hopitals i have

[itex]\frac{1}{e}[/itex] which is less then 1 so ratio test says converges
 
Last edited:
  • #6
Yes, the limit of the ratio is 1/e so it converges.
 
  • #7
Awesome thanks... and I will try to work on form... that was the edited version you should have seen what i had first lol!
 

1. What is L'Hospital's Rule?

L'Hospital's Rule is a mathematical theorem used to evaluate limits of indeterminate forms, where the limit of the numerator and denominator both approach 0 or infinity. It is named after its creator, French mathematician Guillaume de l'Hospital.

2. When should I use L'Hospital's Rule?

L'Hospital's Rule should only be used when evaluating limits of indeterminate forms. This means that the limit cannot be easily determined using traditional methods, such as direct substitution or factoring.

3. How do I apply L'Hospital's Rule?

To apply L'Hospital's Rule, take the derivative of both the numerator and denominator of the fraction in the limit. Then, evaluate the limit of the resulting fraction. If it still results in an indeterminate form, repeat the process until the limit can be determined.

4. What are the indeterminate forms that L'Hospital's Rule can evaluate?

L'Hospital's Rule can evaluate the indeterminate forms 0/0 and infinity/infinity. It can also be extended to evaluate other indeterminate forms, such as infinity - infinity and 0 x infinity.

5. Are there any limitations to using L'Hospital's Rule?

Yes, there are some limitations to using L'Hospital's Rule. It should only be used to evaluate limits of indeterminate forms, and should not be used as a shortcut for finding limits in all cases. It also cannot be used to evaluate limits of sequences or series.

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