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.

It's well-known that $\lim_{x\rightarrow\infty}\left[1+\frac{a}{x}\right]^x=\operatorname{exp}[a]$. I am wondering how fast does the limit converge as $x$ increases, and how the speed of convergence depends on $a$. That is, I would like to find out what $f(x,a)$ is where:

$$\left|\left[1+\frac{a}{x}\right]^x-\operatorname{exp}[a]\right|=\mathcal{O}(f(x,a))$$

However, I'm having trouble evaluating that absolute value. Any tips? Perhaps this a known result...

share|improve this question
    
Take logarithms, set $h = 1/x$, and turn it into a question about the rate of convergence of $\frac{\log 1+ha)}{h} - a$. This last expression is $O(ha^2/2)$, so your $f$ is something like $f(x,a) = \frac{a^2}{2x}$. –  Hans Engler Dec 15 '12 at 22:31
add comment

1 Answer

up vote 3 down vote accepted

$$\left(1+\frac{a}{x}\right)^x=\exp\left(x\ln\left(1+\frac{a}{x}\right)\right).$$ When $x\gg a$, expand the logarithm in a series, giving $$\left(1+\frac{a}{x}\right)^x\approx\exp\left(x\left[\frac{a}{x}-\frac{a^2}{2x^2}\right]\right)=e^a\cdot e^{-a^2/2x}\approx e^a\left(1-\frac{a^2}{2x}\right).$$ So it seems the answer to your question is $$f(x,a)=\frac{a^2 e^a}{2x}.$$

share|improve this answer
    
Thanks! Makes complete sense (and I keep forgetting the "exponentiating the log" trick)... –  M.B.M. Dec 16 '12 at 4:34
add comment

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.