Determining Series Convergence using the Ratio Test

  • Thread starter BraedenP
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  • #1
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Homework Statement



I'm asked to specifically use the Ratio Test (formula below) to determine whether this series converges or diverges (if it converges, the value to which it converges is not needed.)

[tex]\sum_{n=1}^{\infty}\frac{n}{(e^n)^2}[/tex]

Homework Equations



Ratio Test:
If [itex]a_n[/itex] is a sequence of positive terms, then the following conditions are true for the limit:
[tex]\lim_{n\to\infty}\frac{a_{n+1}}{a_n}[/tex]
If lim<1, the series converges.
If lim>1, the series diverges.
If lim=1, inconclusive.

The Attempt at a Solution



By simply plugging values into the formula and rearranging the resulting fractions, I get this form:

[tex]\frac{(e^n)^2(n+1)}{(e^{(n+1)})^2\cdot n}[/tex]

From what I can see, applying L'Hopital's rule won't help get this limit into a solvable form, and no algebraic transformations pop out at me as being helpful.

How would one start solving this limit?
 
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Answers and Replies

  • #2
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[tex]
\frac{e^{n^2}}{e^{(n + 1)^2}} = e^{-(n + 1)^2 + n^2} = e^{-2n - 1}
[/tex]
 
  • #3
2,571
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[tex]
\frac{e^{n^2}}{e^{(n + 1)^2}} = e^{-(n + 1)^2 + n^2} = e^{-2n - 1}
[/tex]
And to make that even more explicit

[tex]e^{-(2n + 1)}[/tex]
 
  • #4
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Hmm.. I tried that method, but the limit evaluates to 0, which doesn't help when identifying convergence.

I could try a bunch of other tests on this series, but the assignment specifically says to use the Ratio Test. Any suggestions? Perhaps I could just say that I performed the Ratio Test, but it was inconclusive, so I used another method.
 
  • #5
2,967
5
Hmm.. I tried that method, but the limit evaluates to 0, which doesn't help when identifying convergence.
ORLY? Care to tell us what ratio test you are using?
 
  • #6
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ORLY? Care to tell us what ratio test you are using?
I just factored out the term provided above, and got:

[tex]e^{-(2n+1)}\cdot(\frac{n+1}{n})[/tex]
[tex]\frac{1+\frac{1}{n}}{e^{(2n+1)}}[/tex]

And that approaches 0 as [itex]n\to\infty[/itex]: [itex]\frac{1}{\infty}[/itex].
 
  • #7
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So, how is this inconclusive?
 
  • #8
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So, how is this inconclusive?
Because according to the Ratio Test, if the limit in this form equates to 0, then I can't say anything regarding the convergence of the series.
 
  • #9
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And when could you say that it converges or diverges?
 
  • #10
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And when could you say that it converges or diverges?
I wrote the conditions of the Ratio Test in the OP. If the limit is greater than 1, we can say that it diverges. If it's less than 1, the series converges.
 
  • #11
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So, is zero greater than or smaller than one?
 
  • #12
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Because according to the Ratio Test, if the limit in this form equates to 0, then I can't say anything regarding the convergence of the series.
I think you mean "if the limit in this form equates to 1" then it's inconclusive.
 
  • #13
96
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So, is zero greater than or smaller than one?
OOPS! I memorized the theorem wrong! If it EQUALS 1, the test is inconclusive. Never mind.

Sometimes I wonder how I'm still alive after these retarded mistakes.

Thanks! You're a life saver :)

Edit:


I think you mean "if the limit in this form equates to 1" then it's inconclusive.
Yeah, I figured that out a little too late. :P Thanks :)
 
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