# absolutely convergent, conditionally convergent or divergent

• Aug 4th 2009, 12:39 PM
acosta0809
absolutely convergent, conditionally convergent or divergent
the question is: determine whether the series is absolutely convergent, conditionally convergent or divergent.

$\displaystyle \sum_{n = 1}^{\infty} (-3)^n \frac{n+1}{\exp(n)}$

I found that if i do the ratio test it diverges but im not sure how to do the alternating test, is a just

$\displaystyle \frac{n+1}{\exp(n)}$

or does it include the 3?
• Aug 4th 2009, 01:05 PM
skeeter
Quote:

Originally Posted by acosta0809
the question is: determine whether the series is absolutely convergent, conditionally convergent or divergent.

$\displaystyle \sum_{n = 1}^{\infty} (-3)^n \frac{n+1}{\exp(n)}$

I found that if i do the ratio test it diverges but im not sure how to do the alternating test, is a just

$\displaystyle \frac{n+1}{\exp(n)}$

or does it include the 3?

ratio test ...

$\displaystyle \lim_{n \to \infty} \left| \frac{(-3)^{n+1}(n+2)}{e^{n+1}} \cdot \frac{e^n}{(-3)^n (n+1)}\right|$

$\displaystyle \lim_{n \to \infty} \left| \frac{(-3)(n+2)}{e(n+1)}\right|$

$\displaystyle \left|\frac{-3}{e}\right| \lim_{n \to \infty} \frac{n+2}{n+1}$

$\displaystyle \left|\frac{-3}{e}\right| \cdot 1 > 1$

series diverges
• Aug 4th 2009, 01:05 PM
Krizalid
For each $\displaystyle n\ge1$ it's $\displaystyle \frac{3^{n}(n+1)}{e^{n}}>n+1,$ so this is far to converge absolutely.