- #1
shamieh
- 539
- 0
determine if series is absolutely convergent, conditionally convergent, or divergent
\(\displaystyle \sum^{\infty}_{n = 1} (n^2 + 9)(-2)^{1-n} \)
which i turned into \(\displaystyle \sum^{\infty}_{n = 1} (n^2 + 9)(-2)^{-n+1} \)
so using the ratio test I got:
\(\displaystyle \frac{((n+1)^2 + 9)(-2)^{-n})}{n^2 + 9 * (-2)^{1-n}}\)
which ended up as n--> infinity becoming \(\displaystyle \frac{2}{3}\) therefore by ratio test L < 1 so the series converges
\(\displaystyle \sum^{\infty}_{n = 1} (n^2 + 9)(-2)^{1-n} \)
which i turned into \(\displaystyle \sum^{\infty}_{n = 1} (n^2 + 9)(-2)^{-n+1} \)
so using the ratio test I got:
\(\displaystyle \frac{((n+1)^2 + 9)(-2)^{-n})}{n^2 + 9 * (-2)^{1-n}}\)
which ended up as n--> infinity becoming \(\displaystyle \frac{2}{3}\) therefore by ratio test L < 1 so the series converges