Can't find the expression for comparision test on a serie

1. Jan 11, 2013

joao_pimentel

Hi guys

I'm trying to find out the nature of this serie

$$\sum_{n=1}^{\infty}\left[ e^{1/n}-e^{-1/n}\right]$$

I know the ratio and root tests are inconclusive, just the comparision test works (it diverges)

But with which expression shall I compare?

Thank you so much

2. Jan 11, 2013

tiny-tim

hi joao_pimentel!

have you tried ∑ 2/n ?

(yes, i know it doesn't work, but why doesn't it work, and how might you fix it? )

3. Jan 11, 2013

joao_pimentel

Hi

Thanks for the reply :)

I'm trying to find this limit

$$\lim\frac{e^{1/n}-e^{-1/n}}{\frac{2}{n}}$$

let's see what can I get...

4. Jan 11, 2013

joao_pimentel

ok, the limit is zero

$$\lim\frac{e^{1/n}-e^{-1/n}}{\frac{2}{n}}=\frac{e^{0}-e^{0}}{\frac{2}{0}}=0$$

humm, what changes shall I make?!?

Last edited: Jan 11, 2013
5. Jan 11, 2013

tiny-tim

no, i mean is e1/n - e-1/n greater or less than 2/n ?

and by roughly how much?

6. Jan 11, 2013

joao_pimentel

it doesn't work because 2/n>e^(1/n)-e^(-1/n)

and how can I fix it? :)

7. Jan 11, 2013

tiny-tim

that's right, it's less

(and you need it to be greater than something, to prove it diverges)

but roughly how much less is it?

8. Jan 11, 2013

joao_pimentel

humm, intresting, I'm realising that both expression are really close as n goes inf.

ok, my last limit was incorrect, sorry

$$\lim\frac{e^{1/n}-e^{-1/n}}{\frac{2}{n}}=\frac{e^{0}-e^{0}}{\frac{2}{\infty}}=\frac{1-1}{0}=\frac{0}{0}=Ind$$

and solving it it gives 1

9. Jan 11, 2013

joao_pimentel

Thank you tiny-tim

of course, I may compare it with 1/x :)

10. Jan 12, 2013

joao_pimentel

By the way, how did you find out that 2/n and the other expression are very similar as n goes to inf?

By the Mc-Lauren expansion serie?

Thank you

11. Jan 12, 2013

tiny-tim

yes of course

12. Jan 13, 2013

joao_pimentel

Thank you so very much again :)

you've been most helpful

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