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I need to use the ratio test to determine if the infinite series $\displaystyle \frac{3^n}{2^n +1}$ converges or diverges.

I began with $\frac{a_{n+1}}{a_n}$ and got to $\frac{3(2^n+1)}{2^{n+1}+1}$ but I don't really know where to go next...

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    $3^n/(2^n+1)$ is not an infinite series. Did you mean, the infinite series $\sum_1^{\infty}3^n/(2^n+1)$?2012-11-19

2 Answers 2

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You can also do this which is far simpler.

${3^n\over 2^n + 1} = \left({3\over 2}\right)^n{1\over{1 + 1/2^n}}.$ The second factor converges to 1. The first increases without bound. This diverges.

You would only use the ratio test if you were interested in the convergence of the series

$\sum_n {3^n\over 1+ 2^n}$

This would diverge too since the terms do not go to zero.

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you came to

$\frac{3(2^n+1)}{2^{n+1}+1}$


divide numerator and denominator with ${2^n}$.

$\frac{3(1+\frac{1}{2^n})}{2+\frac{1}{2^{n}}}$

take the limit of $n$ to infinity.

$\lim_{n\to\infty}\frac{3(1+\frac{1}{2^n})}{2+\frac{1}{2^{n}}}=\frac{3}{2}$

this has to diverge, since

$\frac{3}{2}>1$

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    believe it or not, it LaTeX gets easier with practice. It just takes time getting used to it, then it becomes second nature.2012-11-19