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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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phiBest ResponseYou've already chosen the best response.0
I would try summing the negative terms and the positive terms
 one year ago

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

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

ZarkonBest 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
 one year ago

ZarkonBest ResponseYou've already chosen the best response.2
for \[\sum_{n=0}^{\infty} n\left(\frac{1}{ 3 }\right)^n\] write as \[\frac{1}{3}\sum_{n=0}^{\infty} n\left(\frac{1}{ 3 }\right)^{n1}\] 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
 one year ago

ZarkonBest ResponseYou've already chosen the best response.2
I get \[\frac{3}{16}+\frac{3}{2}=\frac{21}{16}\]
 one year ago

frxBest ResponseYou've already chosen the best response.0
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
 one year ago

ZarkonBest ResponseYou've already chosen the best response.2
if a series converges uniformly then it can integrated or differentiated term by term
 one year ago

frxBest ResponseYou've already chosen the best response.0
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!
 one year ago
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