# 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?

Note that:

So:

Then, as:

the series is not convergent.

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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:

- ( b_n ) is positive and decreasing for all ( n ).
- 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 ):

- ( b_n ) is positive since both the numerator and denominator are positive for all ( n ).
- 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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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