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

The question is to find:

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


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up vote 3 down vote accepted

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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