Question

How do you determine the convergence or divergence of `Sigma (-1)^(n+1)cschn` from `[1,oo)`?

Answer 1

The series converges

Explanation:

Perform the ratio test

`a_n=(-1)^(n+1)csch n`

Therefore,

`|a_(n+1)/a_n|=|((-1)^(n+2)csch(n+1))/((-1)^(n+1)csch n)|`

`=|(csch(n+1))/csch n|`

Then,

`lim_(n->oo)|a_(n+1)/a_n|=lim_(n->oo)|(csch(n+1))/csch n|`
c
`=lim_(n->oo)|(e^(n)-1/e^(n))/(e^(n+1)-1/e^(n+1))|`

`=lim_(n->oo)|(((e^n-1/e^n))/(e(e^n-1/e^n))|`

`=1/e`

As,

`lim_(n->oo)|a_(n+1)/a_n|=1/e<1`

We conclude that the series converges ( absolutely) by the ratio test