## anonymous 5 years ago how do you determine if a series is absolutely ,conditionally convergent or divergent for a problem like this sigma (-1)^n/2^n

1. watchmath

The absolute series is a geometric series with ratio 1/2. Hence it is absolute convergent.

2. anonymous

how would you test if it is conditionally convergent

3. anonymous

since terms go to zero and it alternates it is conditionally convergent.

4. anonymous

would you use the latenating test along with another test?

5. anonymous

I believe that this is an alternating series, and is convergent. its absolutely convergent if its still convergent using |(-1)^b/2^n| with absolute value operator otherwise its conditionally convergent

6. anonymous

if it alternates, which this does because of the $(-1)^n$ then all you need for conditional convergence is that the terms go to zero.

7. anonymous

okay so you use the alternating test along with another test to see is it is absolutely or conditionally

8. anonymous

to show absolute convergence you just check $\sum\frac{1}{2^n}$ which as watchmath said, is a geometric series with $r=\frac{1}{2}$

9. anonymous

ok i sense it is not clear from your question. absolute convergence is stronger than conditional. it is converges absolutely then it certainly converges conditionally

10. anonymous

for an alternating series all you have to check for conditional convergence is that the terms go to zero.

11. watchmath

This is an example of series that conditionally convergent $$\sum (-1)^n\frac{1}{n}$$ The absolute value series is divergent since it is a harmonic series. But the original series is convergent by alternating series test. So it is always better to check it for the absolute series first.