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anonymous
 4 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?
anonymous
 4 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?

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phi
 4 years ago
Best ResponseYou've already chosen the best response.0I would try summing the negative terms and the positive terms

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Don't really know what you mean, should I expand the serie a couple of times then sum?

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2\[\sum_{n=0}^{\infty} (1)^{n}\frac{ n+2 }{ 3^{n} }\] \[\sum_{n=0}^{\infty} n\left(\frac{1}{ 3 }\right)^n+2\sum_{n=0}^{\infty} \left(\frac{1}{ 3 }\right)^n\]

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2\[2\sum_{n=0}^{\infty} \left(\frac{1}{ 3 }\right)^n\] is just a geometric sum...should be easy

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2for \[\sum_{n=0}^{\infty} n\left(\frac{1}{ 3 }\right)^n\] write as \[\frac{1}{3}\sum_{n=0}^{\infty} n\left(\frac{1}{ 3 }\right)^{n1}\] then as \[\frac{1}{3}\sum_{n=0}^{\infty} n\left(x\right)^{n1}\] integrate this wrt x then compute the sum... then differentiate and plug in 1/3

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2I get \[\frac{3}{16}+\frac{3}{2}=\frac{21}{16}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Your 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

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2if a series converges uniformly then it can integrated or differentiated term by term

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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!
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