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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.

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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. – S. Snape 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
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!

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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
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
@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

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: $$\lim_{n\to\infty}\left(1-{a_n\over n}\right)^n=\exp\bigl(-e^2\bigr)\ .$$

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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 $a_n=(1+\frac2n)^n$.

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