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

• joao_pimentel
In summary, the conversation is about finding the nature of a series and comparing it with other expressions. The ratio and root tests are inconclusive, but the comparison test shows that the series diverges. The expression ∑ 2/n does not work and cannot be used to prove divergence. The limit is found to be zero and the conversation explores how to fix the expression to prove divergence. The Mc-Lauren expansion series is used to show that the expression is very similar to 2/n as n goes to infinity. The final conclusion is that the series diverges and can be compared with 1/x. The conversation ends with gratitude for the helpful responses.
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

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? )

tiny-tim said:
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

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...

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

and by roughly how much?

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

and how can I fix it? :)

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?

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

Thank you tiny-tim

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

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

joao_pimentel said:
By the Mc-Lauren expansion serie?

yes of course

Thank you so very much again :)

## 1. What is the comparison test for series?

The comparison test for series is a method used to determine whether a given series converges or diverges by comparing it to a known series or sequence with known convergence properties.

## 2. How does the comparison test work?

The comparison test works by comparing the given series to a known series or sequence, typically one that is easier to determine convergence for. If the given series is larger than the known series and the known series diverges, then the given series must also diverge. If the given series is smaller than the known series and the known series converges, then the given series must also converge.

## 3. What is the expression for the comparison test on a series?

The expression for the comparison test on a series is the ratio of the terms in the given series and the known series or sequence. This ratio is used to determine whether the given series is larger or smaller than the known series and thus determine the convergence of the given series.

## 4. What types of series can the comparison test be applied to?

The comparison test can be applied to both infinite and finite series, as long as the terms in the series are positive. It can also be applied to both series with positive and negative terms, as long as the absolute values of the terms are used in the comparison.

## 5. Are there any limitations to the comparison test for series?

Yes, the comparison test may not be applicable if the given series is not comparable to a known series or sequence or if the known series or sequence does not have known convergence properties. In such cases, other convergence tests may need to be used.

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