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Finding the limit below..:

$$\lim_{n \rightarrow +\infty}\frac{n}{2^n}= ?$$

I really think its 0. But intuitively, infinity over infinity. how can that be? indeterminate forms? Thanks

Shuchang
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Keneth Adrian
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  • More than needed: For $a_n=n/2^n$, you can show $\lim\limits_{n\rightarrow\infty}{a_{n+1}\over a_n}={1\over2}<1$. This implies that your limit has value $0$ ($(a_n)$ is eventually dominated by a geometric sequence, as the usual proof of the Ratio Test for series shows.) – David Mitra Aug 29 '13 at 11:18

3 Answers3

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Intuitively, $2^n$ grows much faster than $n$.

Note that by the Binomial Theorem, $2^n=(1+1)^n=1+n+\frac{n(n-1)}{2}+\cdots$.

In particular, if $n\gt 1$, we have $2^n\ge \dfrac{n(n-1)}{2}$.

Thus $0\lt \dfrac{n}{2^n}\le \dfrac{2}{n-1}$. But $\frac{2}{n-1}$ approaches $0$ as $n\to\infty$, so by Squeezing, so does $\dfrac{n}{2^n}$.

Another way: You can use L'Hospital's Rule to show $\lim_{x\to\infty}\frac{x}{2^x}=0$.

André Nicolas
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The limit of form that infinity over infinity can, in most cases, be easily computed using L'Hospital rule.

Thus $$\lim_{n\rightarrow\infty}\frac{n}{2^n}=\lim_{n\rightarrow\infty}\frac{n'}{(2^n)'}=\lim_{n\rightarrow\infty}\frac{1}{2^n\ln 2}=0$$

Shuchang
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The term indeterminate forms applies in your case only if you want to calculate the limit of a sequence of quotients as a quotient of limits: $$\frac{\lim \limits_{n \rightarrow \infty}{n}}{\lim \limits_{n \rightarrow \infty}{2^n}}$$ But this view is not necessary, because you just have a sequence of rational numbers and you can directly apply the $\epsilon$ definition of a limit to your sequence: Given $\epsilon > 0$, you have $0\lt \frac{n}{2^n} \le \frac{2}{n-1} < \epsilon$ for $n > \frac{2}{\epsilon} + 1$, this shows that the sequence converges to 0.

gammatester
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