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

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Homework Help Overview

The discussion revolves around the convergence or divergence of the series Ʃ n/e^n. Participants are exploring methods to analyze this series, particularly focusing on the use of L'Hospital's Rule and the Ratio Test.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss using the integral test and express uncertainty about applying L'Hospital's Rule. There are mentions of potential errors in setting up the integral and the need for clarification on the correct form of the series. Some participants suggest switching to the Ratio Test as an alternative method.

Discussion Status

Some participants have attempted to apply the Ratio Test and have shared their findings, while others are still questioning the setup of the integral and the application of L'Hospital's Rule. There is a mix of approaches being explored, with no explicit consensus reached on the best method to use.

Contextual Notes

Participants note that the original problem statement is correctly identified as n/e^n, and there are discussions about the implications of the infinity-infinity form encountered during analysis. There are also references to feedback received on previous attempts at solving the problem.

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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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)?
 
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
 
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.
 
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'hospital's i have

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

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