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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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Manasa, Let me know if I typeset your question correctly. –  amWhy Nov 19 '12 at 2:17
2  
$3^n/(2^n+1)$ is not an infinite series. Did you mean, the infinite series $\sum_1^{\infty}3^n/(2^n+1)$? –  Gerry Myerson Nov 19 '12 at 2:22

2 Answers 2

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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latex is really hard! –  thkang Nov 19 '12 at 2:32
    
believe it or not, it LaTeX gets easier with practice. It just takes time getting used to it, then it becomes second nature. –  amWhy Nov 19 '12 at 2:55

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