How do you test the alternating series #Sigma (-1)^n(2^(n-2)+1)/(2^(n+3)+5)# from n is #[0,oo)# for convergence?

Question

1518 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-19T00:01:15

Note that:

#a_n = (2^(n-2)+1)/(2^(n+3)+5)#

#a_n = (2^n*2^-2+1)/(2^n*2^3+5)#

#a_n = (2^n(1/4+2^-n))/(2^n(8+5*2^-n))#

#a_n =(1/4+2^-n)/(8+5*2^-n)#

So:

#lim_(n->oo) a_n = 1/32#

Then, as:

#lim_(n->oo) a_n != 0#

the series is not convergent.