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

Emily Ammerman · Accepted Answer

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.

William Abdalla · Answer

undefined To test the convergence of the alternating series $ \sum_{n=0}^{\infty} (-1)^n \frac{2^{n-2} + 1}{2^{n+3} + 5} $ for $ n $ from $ 0 $ to $ \infty $, you can use the Alternating Series Test.

The Alternating Series Test states that if an alternating series $ \sum_{n=0}^{\infty} (-1)^n b_n $ meets the following conditions, then it converges:

1. $ b_n $ is positive and decreasing for all $ n $.
2. The limit of $ b_n $ as $ n $ approaches infinity is 0: $ \lim_{n \to \infty} b_n = 0 $.

First, let's analyze the given series:

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

Now, check if $ b_n $ is positive and decreasing for all $ n $:

1. $ b_n $ is positive since both the numerator and denominator are positive for all $ n $.
2. To check if $ b_n $ is decreasing, find $ b_{n+1} - b_n $ and verify if it's negative:

$$ b_{n+1} - b_n = \frac{2^{(n+1)-2} + 1}{2^{(n+1)+3} + 5} - \frac{2^{n-2} + 1}{2^{n+3} + 5} $$

Simplify this expression and check if it's negative for all $ n $. If it is, then $ b_n $ is decreasing.

After confirming that $ b_n $ is positive and decreasing, the next step is to check the limit as $ n $ approaches infinity:

$$ \lim_{n \to \infty} b_n = \lim_{n \to \infty} \frac{2^{n-2} + 1}{2^{n+3} + 5} $$

Solve this limit and check if it equals 0. If the limit is indeed 0, then the alternating series $ \sum_{n=0}^{\infty} (-1)^n b_n $ converges by the Alternating Series Test.