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

1 Answer
Nov 7, 2017

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