# 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?

Oct 19, 2017

Note that:

${a}_{n} = \frac{{2}^{n - 2} + 1}{{2}^{n + 3} + 5}$

${a}_{n} = \frac{{2}^{n} \cdot {2}^{-} 2 + 1}{{2}^{n} \cdot {2}^{3} + 5}$

${a}_{n} = \frac{{2}^{n} \left(\frac{1}{4} + {2}^{-} n\right)}{{2}^{n} \left(8 + 5 \cdot {2}^{-} n\right)}$

${a}_{n} = \frac{\frac{1}{4} + {2}^{-} n}{8 + 5 \cdot {2}^{-} n}$

So:

${\lim}_{n \to \infty} {a}_{n} = \frac{1}{32}$

Then, as:

${\lim}_{n \to \infty} {a}_{n} \ne 0$

the series is not convergent.