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This is a question from my homework. Please help me!

The question is to find:

$\lim_{n\to\infty} \frac{2^{n+1}}{ 7^n}$

Thanks

2 Answers 2

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Hint: Use this fact: $ \lim_{n \to +\infty} a^n = 0 $ if $|a| < 1$ (proof?). Also, notice $\frac{2^{n+1}}{7^n} = 2\left(\frac{2}{7}\right)^n $

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    i still cannot accept your answer, it keeps saying that "you can accept an answer in 9 minutes", but 9 minutes have passed away so long ago2012-10-28
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Set $a_n=\frac{2^{n+1}}{7^n}$ and consider the series $\sum_{n=0}^{\infty}a_n$. Since $\lim_{n\rightarrow\infty}\sqrt[n]{a_n}=\frac{2}{7}<1$, by root test we obtain that the series is convergent. Hence general term of the series goes to zero, i.e,

$\lim_{n\rightarrow \infty}\frac{2^{n+1}}{7^n}=0$.