Question

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?

Answer 1

@phi

Answer 2

I would try summing the negative terms
and the positive terms

Answer 3

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

Answer 4

\[\sum_{n=0}^{\infty} (-1)^{n}\frac{ n+2 }{ 3^{n} }\]

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

Answer 5

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

is just a geometric sum...should be easy

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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!

Answer 11

good