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

[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

 

[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? )

 

[joao_pimentel]

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? )

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...

 

[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?!?

 

[tiny-tim]

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

and by roughly how much? ​

 

[joao_pimentel]

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

and how can I fix it? :)

 

[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? ​

 

[joao_pimentel]

humm, interesting, 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

 

[joao_pimentel]

Thank you tiny-tim

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

 

[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

 

[tiny-tim]

By the Mc-Lauren expansion serie?

yes of course

 

[joao_pimentel]

Thank you so very much again :)

you've been most helpful