## frx 2 years ago Is the serie $\sum_{n=0}^{\infty} (-1)^{n}\frac{ n+2 }{ 3^{n} }$ convergent or divergent? Calculate is't value. $\lim_{n \rightarrow \infty} \left| \left( \frac{ (-1)^{n+1}(n+3) }{ 3^{n+1} } \right)\left( \frac{ 3^{n} }{ (-1)^{n}(n+2) } \right) \right|$ $\frac{ 1 }{ 3 } \lim_{n \rightarrow \infty} \left| \frac{ n+3 }{n+2 } \right|=\frac{ 1 }{ 3 }$ $\frac{ 1 }{ 3 } < 1; Convergent$ But how do I calculate it's value?

1. frx

@phi

2. phi

I would try summing the negative terms and the positive terms

3. frx

Don't really know what you mean, should I expand the serie a couple of times then sum?

4. Zarkon

$\sum_{n=0}^{\infty} (-1)^{n}\frac{ n+2 }{ 3^{n} }$ $\sum_{n=0}^{\infty} n\left(\frac{-1}{ 3 }\right)^n+2\sum_{n=0}^{\infty} \left(\frac{-1}{ 3 }\right)^n$

5. Zarkon

$2\sum_{n=0}^{\infty} \left(\frac{-1}{ 3 }\right)^n$ is just a geometric sum...should be easy

6. Zarkon

for $\sum_{n=0}^{\infty} n\left(\frac{-1}{ 3 }\right)^n$ write as $\frac{-1}{3}\sum_{n=0}^{\infty} n\left(\frac{-1}{ 3 }\right)^{n-1}$ then as $\frac{-1}{3}\sum_{n=0}^{\infty} n\left(x\right)^{n-1}$ integrate this wrt x then compute the sum... then differentiate and plug in -1/3

7. Zarkon

I get $\frac{-3}{16}+\frac{3}{2}=\frac{21}{16}$

8. frx

Your answer is correct according to my key, what's the metod you're using named? I think I need to study the concept a bit closer

9. Zarkon

if a series converges uniformly then it can integrated or differentiated term by term

10. frx

I think I've found the section in my book now, don't get the concept yet but now I know where to start working on it. Thank you!

11. Zarkon

good