1. ## more sequence/series

from n=1 to infinity of

((-1)^n * n)/3^n
does it converge or diverge, and what is the sum?
good luck!!!

2. It's convergent by the alternating series test.

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

$\displaystyle \lim_{n\rightarrow{\infty}}\frac{n}{3^n}=0$ while $\displaystyle a_{n+1}\leq{a_n}$.

3. Hello, alex_nicole!

$\displaystyle \sum^{\infty}_{n=1} (\text{-}1)^n\,\frac{n}{3^n}$

We have: . . . . $\displaystyle S \;=\;-\frac{1}{3} + \frac{2}{3^2} - \frac{3}{3^3} + \frac{4}{3^4} + \cdots$

Multiply by $\displaystyle \frac{1}{3}\!:\;\;\frac{1}{3}S \;=\qquad-\frac{1}{3^2} + \frac{2}{3^3} - \frac{3}{3^4} + \cdots$

Add:. . . . . . . $\displaystyle \frac{4}{3}S \;=\;-\frac{1}{3} + \frac{1}{3^2} - \frac{1}{3^3} + \frac{1}{3^4} + \cdots$

$\displaystyle \text{We have: }\qquad\quad \frac{4}{3}S \;=\;-\frac{1}{3}\underbrace{\left(1 - \frac{1}{3} + \frac{1}{3^2} - \frac{1}{3^2} + \cdots\right)}_{\text{gemetric seres}}$

. . The series has the sum: .$\displaystyle \frac{1}{1-\left(-\frac{1}{3}\right)} \:=\:\frac{3}{4}$

And we have: . $\displaystyle \frac{4}{3}S \:=\:-\frac{1}{3}\left(\frac{3}{4}\right) \:=\:-\frac{1}{4}$

Therefore: . $\displaystyle S \;=\;\frac{3}{4}\left(-\frac{1}{4}\right) \;=\;\boxed{-\frac{3}{16}}$