Can alternating series be grouped as geometric series for convergence?

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The discussion centers on the convergence of the alternating series \(\sum(-1)^n(\frac{3}{2})^n\). It is established that this series fails the alternating series test because the limit of \(a_n = (\frac{3}{2})^n\) does not approach zero as \(n\) approaches infinity. Consequently, the series diverges, as the absolute value of the common ratio \(|r| = \frac{3}{2}\) exceeds 1. The consensus is that regardless of the alternating nature, if the summand does not approach zero, the series cannot converge.

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I'm reviewing power series for use in differential equations and I'm having some trouble remembering how to deal with alternating series.

For instance, if I have:
\sum(-1)^n(\frac{3}{2})^n

if a_n=(\frac{3}{2})^n
This fails the alternate series test because the limit of a_n as n goes to infinity doesn't equal 0.

Can I group the (-1)^n into the fraction and call it a geometric series? In that case, it would diverge, |r| would be greater then 1.
 
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kdinser said:
I'm reviewing power series for use in differential equations and I'm having some trouble remembering how to deal with alternating series.

For instance, if I have:
\sum(-1)^n(\frac{3}{2})^n

if a_n=(\frac{3}{2})^n
This fails the alternate series test because the limit of a_n as n goes to infinity doesn't equal 0.

Can I group the (-1)^n into the fraction and call it a geometric series? In that case, it would diverge, |r| would be greater then 1.

Yes, both ways are fine.
 
thanks for the help.
 
If the summand doesn't go to zero, the series cannot converge, regardless of whether it is alternating or not (using the most common definition of convergence).
 
Data said:
If the summand doesn't go to zero, the series cannot converge, regardless of whether it is alternating or not (using the most common definition of convergence).

Yes, you're right. This is the best way to solve the problem. The summand doesn't go to zero so the series diverges.
 

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