- #1
shamieh
- 538
- 0
determine if series is absolutely convergent, conditionally convergent, or divergent
$$\sum^{\infty}_{n = 1} (n^2 + 9)(-2)^{1-n} $$
which i turned into $$\sum^{\infty}_{n = 1} (n^2 + 9)(-2)^{-n+1} $$
so using the ratio test I got:
$$\frac{((n+1)^2 + 9)(-2)^{-n})}{n^2 + 9 * (-2)^{1-n}}$$
which ended up as n--> infinity becoming $$\frac{2}{3}$$ therefore by ratio test L < 1 so the series converges
$$\sum^{\infty}_{n = 1} (n^2 + 9)(-2)^{1-n} $$
which i turned into $$\sum^{\infty}_{n = 1} (n^2 + 9)(-2)^{-n+1} $$
so using the ratio test I got:
$$\frac{((n+1)^2 + 9)(-2)^{-n})}{n^2 + 9 * (-2)^{1-n}}$$
which ended up as n--> infinity becoming $$\frac{2}{3}$$ therefore by ratio test L < 1 so the series converges
