# Limit of sequence

The question is to find:

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

Thanks

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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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thanks, but i do not know why i cannot accept your answer!!! –  Jaden Q Oct 28 '12 at 9:09
i think you need to wait couple of minutes.. –  ILoveMath Oct 28 '12 at 9:10
i think now you can accept the answer –  ILoveMath Oct 28 '12 at 9:34
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 ago –  Jaden Q Oct 28 '12 at 9:43

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,

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

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