Series: test (-3)^(n+1) / 2^(3n) for convergence

[flakine]

Test the series for convergence or divergence:

n=1, as n -> infinity, summation: (-3)^(n+1)/2^(3n)

Would it be correct to say this series converge, being a geometric series with r = (-3/8), so |r| < 1? Or is it invalid to use the geometric series for negetives?

 

[galactus]

Do you just have to test?. The ratio test is good for these types.

If p is less than 1, it converges.

\(\displaystyle \L\\p=\frac{(-3)^{n+2}}{2^{3(n+1)}}\cdot\frac{2^{3n}}{(-3)^{n+1}}\)

Try the geometric series formula. Be careful. (-3)^(n+1) is different than
-3^(n+1)

To actually find what it converges to:

\(\displaystyle \L\\\sum_{n=1}^{\infty}\frac{(-3)^{n+1}}{2^{3n}}={-}3\sum_{n=1}^{\infty}(\frac{-3}{8})^{n}\)

 

[pka]

We can also note that:
\(\displaystyle \L\frac{{\left( { - 3} \right)^{n + 1} }}{{2^{\left( {3n} \right)} }} = \frac{{\left( { - 1} \right)^{n + 1} \left( 3 \right)^{n + 1} }}{{8^n }}3\left( { - 1} \right)^{n + 1} \left( {\frac{3}{8}} \right)^n\)

Thus we have an alternating series and the sequence \(\displaystyle 3\left( {\frac{3}{8}} \right)^n\) is a decreasing, null sequence.
Therefore, by the alternating series test the series converges