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?

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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.

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