- #1

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- Homework Statement
- Use the ratio test for the series ##\sum_{n=1}^{\infty} b_{n}## where ##b_{1} = 5## and ##b_{n}= \frac {(-1)^{n}nb_{n-1}}{3n+1}## for ##n \geq 2##

- Relevant Equations
- ratio test: ##L = \lim_{n \to \infty} \lvert \frac {a_{n+1}}{a_{n}} \rvert## will converge if ##L < 1## and will diverge if ##L > 1##

So I am having some difficulty expressing this series explicitly. I just tried finding some terms

##b_{0} = 5##

I am assuming I am allowed to use that for ##b_{1}## for the series, even if the series begins at ##n=1##? With that assumption, I have

##b_{1} = -\frac {5}{4}##

##b_{2} = - \frac{5}{14}##

##b_{3} = \frac {3}{28}##

## b_{4} = \frac {3}{91}##

I am not seeing a pattern. I also alternatively tried using the ratio test not in explicit terms

##\lim_{n \to \infty} \lvert \frac {(-1)^{n+1}(n+1)b_{n}}{3(n+1)+1} \cdot \frac {3n+1}{(-1)^{n}nb_{n-1}} \rvert## and after substituting I ended up with ##\lvert \frac{b_{n}}{b_{n-1}} \rvert## which to me is inconclusive.

##b_{0} = 5##

I am assuming I am allowed to use that for ##b_{1}## for the series, even if the series begins at ##n=1##? With that assumption, I have

##b_{1} = -\frac {5}{4}##

##b_{2} = - \frac{5}{14}##

##b_{3} = \frac {3}{28}##

## b_{4} = \frac {3}{91}##

I am not seeing a pattern. I also alternatively tried using the ratio test not in explicit terms

##\lim_{n \to \infty} \lvert \frac {(-1)^{n+1}(n+1)b_{n}}{3(n+1)+1} \cdot \frac {3n+1}{(-1)^{n}nb_{n-1}} \rvert## and after substituting I ended up with ##\lvert \frac{b_{n}}{b_{n-1}} \rvert## which to me is inconclusive.

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