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.

If $a_n=(1+\frac{2}{n})^n$ , then find $$\lim_{n \to \infty}(1-\frac{a_n}{n})^n$$.

Trial: Can I use $$\lim_{n \to \infty}a_n=e^2$$ Again $$\lim_{n \to \infty}(1-\frac{a_n}{n})^n=\exp(-e^2)$$ Please help.

share|improve this question
1  
No this is not rigorous –  Amr Jan 21 '13 at 12:05
    
Just to reiterate: when n is 10^20, a_n is very close to $e^2$, and $a_n / n$ is relatively close to $e^2 / n$. But then taking to the power of n screws everything up. –  Adam Rubinson Jan 21 '13 at 12:33
    
You can look at the definition(s) of e and find the exact difference between $a_n$ and $e^2$ in terms of n. Enjoy –  Adam Rubinson Jan 21 '13 at 12:37
    
@AdamRubinson: Aren't the OP using this wrong assumption that the composition of two convergent sequence is again a convergent seq.?? However, I am thinking on the problem without that trial. –  B. S. Jan 21 '13 at 12:40
    
Hold on, is it true that (1+2/n)^2 and (1+1/n)^2n both tend to e^2? –  Adam Rubinson Jan 21 '13 at 12:45
add comment

3 Answers

up vote 2 down vote accepted

Due to $$(1-\frac{a_n}{n})^n=\left[\left(1-\frac{a_n}{n}\right)^{\frac{n}{-a_n}}\right]^{\frac{-a_n}{n}n}=\left[\left(1-\frac{a_n}{n}\right)^{\frac{n}{-a_n}}\right]^{-a_n}.$$ $$\lim_{n\to\infty}\left(1-\frac{a_n}{n}\right)^{\frac{n}{-a_n}}=e$$ and $$\lim_{n\to \infty}(-a_n)=-e^2$$

Let $A_n=\left(1-\frac{a_n}{n}\right)^{\frac{n}{-a_n}}$ and $B_n=-a_n$, by the "claim" below, you can get the result!

share|improve this answer
2  
Could you elaborate? –  Michael Albanese Jan 21 '13 at 12:51
    
I think this is nonsense. Edit: I wrote this before Riemann edited his answer –  Adam Rubinson Jan 21 '13 at 12:59
    
Here is a claim says that: If $\lim\limits_{n\to \infty}a_n=a>0$ and $\lim\limits_{n\to \infty}b_n=b$, then $$\lim_{n\to\infty}a_n^{b_n}=a^b.$$ –  Riemann Jan 21 '13 at 13:04
1  
To be honest, I am Chinese working in university(NWU), and this question is an easy problem for Chinese students. There is nothing awesome!! –  Riemann Jan 21 '13 at 13:19
1  
@Belgi $a_n=\left(1+\frac{2}{n}\right)^n=\left[\left(1+\frac{2}{n}\right)^{\frac{n}{2}}‌​\right]^2$, and this implies that $a_n\to e^2.$ Ok????? –  Riemann Jan 21 '13 at 14:12
show 7 more comments

This is a very nice problem; but it does not file under "composite limits", i.e., limits of the form $\lim_{x\to \xi} f\bigl(g(x)\bigr)$.

It is agreed that ${\displaystyle\lim_{n\to\infty} a_n=\lim_{n\to\infty}\left(1+{2\over n}\right)^n = e^2}$.

The function $u\mapsto {\rm Log}(1-u)$ is analytic for $|u|<1$. Therefore "by Taylor's Theorem" there is a function $u\mapsto g(u)$, analytic for $|u|<1$, such that $$\log(1-u)=-u + u^2 g(u)\ .\qquad(1)$$ It follows that there is an $M>0$ such that $$\bigl|g(u)\bigr|\leq M\qquad\left(|u|\leq{1\over2}\right)\ .$$ If $x$ and $y$ are arbitrary real numbers with $0<|xy|<1$ then we get from $(1)$ that $${\log(1-x y)\over x}=-y + x y^2 g(xy)\ .$$ Now put $x:={1\over n}$, $\ y:=a_n$. Then the last equation says $$n\ \log\left(1-{a_n\over n}\right)=-a_n +{1\over n} a_n^2\> g\left({a_n\over n}\right)\ ,\qquad(2)$$ whereby this holds as soon as $n$ is large enough to make $|a_n|<n$.

Now let $n\to\infty$. Then $a_n\to e^2$ and $\ {a_n\over n}\to 0$. Therefore the RHS of $(2)$ converges to $-e^2$. So does the LHS, and we get what we have conjectured all along: $$\lim_{n\to\infty}\left(1-{a_n\over n}\right)^n=\exp\bigl(-e^2\bigr)\ .$$

share|improve this answer
add comment

If $f_n(x)\xrightarrow[n\to\infty]{}f(x)$ uniformly on $E$, $f$ is continuous at $a\in E$ and $a_n\xrightarrow[n\to\infty]{}a$ with $a_n\in E \ \ \forall \ n \in \mathbb N$, then $$f_n(a_n)\xrightarrow[n\to\infty]{}f(a).$$

Indeed, let $\epsilon>0$ and $\delta>0$ such that $|f(x)-f(a)|<\dfrac{\epsilon}{2}$ whenever $|x-a|<\delta$. Let $n_0\in\mathbb N$ such that $|f_n(x)-f(x)|<\dfrac{\epsilon}{2} \text{ and } |a_n-a|<\delta, \ \forall x\in E, \ n\geq n_0$.
Then for $n\geq n_0$: $$ |f_n(a_n)-f(a)|\leq |f_n(a_n)-f(a_n)|+|f(a_n)-f(a)|<\epsilon \ . $$


Now apply the previous to $f_n(x)=(1-\frac xn)^n$ (see this) and $a_n=(1+\frac2n)^n$.

share|improve this answer
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.