Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do you prove $\lim_{n\rightarrow\infty} n(2^{1/n} - 1) = \log 2$ ?

Background: in computer science, if you allocate CPU time to $n$ processes by rate-monotonic scheduling, all the processes get sufficient amount of time when the quantity $U$ called CPU utilization is less than or equal to $n(2^{1/n} - 1)$. It is monotonously decreasing and tends to $\log 2$ when $n\rightarrow\infty$, so if $U \le \log 2$, you can be sure all the processes will be given sufficient CPU time.

share|improve this question

3 Answers 3

up vote 2 down vote accepted

We have $$\lim_{n\to\infty} n(2^{1/n} - 1) = \lim_{t\downarrow 0}{2^t - 1\over t}.$$ Put $f(t) = 2^t$; then the limit is just $f'(0)$. Since $f'(x) = 2^x\log(2)$, we are done.

share|improve this answer

Depends on your knowledge, e.g. with Taylor expansion you have $2^x-1\sim x\log2$ for $x\to 0$, so $$ \lim\limits_{n\to\infty}n(\sqrt[n]{2}-1) = \lim\limits_{x\to 0}\frac1x(2^x-1) = \lim\limits_{x\to 0}\frac1x(x\log2) =\log 2. $$

share|improve this answer

Let $h=1/n$. It is enough to show that $$\lim_{h\to 0}\frac{2^h-1}{h}=\log 2.$$

This just says that the derivative of $2^x$ at $x=0$ is $\log 2$, which is easy to check.

Remark: Equivalently, let $x=1/n$, rewrite our expression as $\frac{e^{(\ln 2) x}}{x}$, and find the limit of this as $x\to 0$ using L'Hospital's Rule. But reducing to the definition of the derivative has a more "elementary" feel.

The mathematically natural approach is the one by Ilya. The Taylor expansion of $2^x$, that is, of $e^{(\ln 2)x}$, tells us about the "local" behaviour of $2^x$ near $x=0$, so it is the right tool to use. The derivative also tells us about local behaviour, but the Taylor expansion is finer-grained.

share|improve this answer
    
Great! I was thinking that there is a more popular word for the 1st order Taylor's expansion (like derivative or a differential), but the lack of coffee didn't allow me to remember that :) I wonder, by the way - if there is a natural way to prove this limit just based on sequences, and not on analysis involving limits of functions. –  Ilya May 30 '12 at 12:58
    
There are things one could do using the binomial expansion, the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$ pops out. More fundamental in a sense, but also a lot more work. –  André Nicolas May 30 '12 at 13:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.