1
$\begingroup$

I have this equation:

$$\lim_{n\to \infty}\left(1+{x\over n} \right)^n=e^x$$

I need to Taylor expand both sides to prove that they are equal. Now I know how to do this using l'Hopital's rule, but I'm not supposed to prove this in that way. This is what I've tried so far:

$$\lim_{n\to \infty}\left(1+{x\over n} \right)^n$$ $$=\lim_{n\to \infty}e^{n\big[\log\big( 1+{x\over n}\big)\big]}$$ $$=\lim_{n\to \infty}\big(1+n\log\big(1+{x\over n}\big)+{n^2\over 2!}{\log^2\big(1+{x\over n}}\big)+O(x^3)\big)$$

This doesn't look anything like the expansion for $e^x$, so I tried to pass the limit through the exponential, expand $\log_{n\to \infty}$, and solve it that way:

$$e^{\lim_{n\to \infty}\big(1+{x\over n} \big)^n}$$ $$e^{\lim_{n\to \infty}\big[1+n\big({x\over n}-{x^2\over 2n^2}+{x^3\over 3n^3}-O(x^4)\big) \big]}$$

Distributing the $n$ into the parentheses, and taking the limit of my result was this:

$$e^{\lim_{n\to \infty}\big(1+{x\over n} \big)^n} = e^{1+x}=e^1e^x$$

And it looks very close to what I need it to be. I also have the problem that there aren't any factorials in the denominators of my expansion. I found this stack exchange question: Limit Question involving logarithmic taylor expansion. However, it didn't help me very much.

How should I expand this equation and what am I doing wrong?

$\endgroup$
  • 1
    $\begingroup$ I think the first approach would be circular since you'd need to compute $n\ln(1+x/n)$ to compute $(1+x/n)^n$. Also, shouldn't your second approach lead to $e^{(e^x)}$?. Have you considered expanding $(1+x/n)^n$ with the binomial theorem? Isn't that how this is normally done? $\endgroup$ – Jam Apr 16 '18 at 22:21
1
$\begingroup$

In fact with taylor expansion of $\ln(1+x)$ at $x=0$ you have $$ \ln(1+x)=x-\frac{x^2}{2}+o\left(x^2\right) $$ So with $\frac{x}{n}\underset{n \rightarrow +\infty}{\rightarrow}0$ you can use it and

$$e^{n\ln\left(1+\frac{x}{n}\right)}=e^{n\left(\frac{x}{n}+o\left(\frac{1}{n}\right)\right)}=e^{x+o\left(1\right)}\underset{n \rightarrow +\infty}{\rightarrow}e^{x}$$

$\endgroup$
2
$\begingroup$

By the binomial theorem, $$ \left( 1 + \frac{x}{n} \right)^n = \sum_{k=0}^n \frac{1}{n^k}\binom{n}{k} x^k = \sum_{k=0}^{\infty} \frac{1}{n^k}\binom{n}{k} x^k, $$ since the binomial coefficient is zero for $k>n$.

Meanwhile, the Taylor expansion of $e^x$ is supposed to be $$ \sum_{k=0}^{\infty} \frac{1}{k!}x^k. $$ Equating powers of $x$, we want to show that $$ \lim_{n \to \infty} \frac{1}{n^k} \binom{n}{k} = \frac{1}{k!}, $$ or expanding out the definition of the binomial coefficient, $$ \lim_{n \to \infty} \frac{n(n-1)\dotsm (n-k+1)}{n^k} = 1. $$ But this is straightforward: $$ \frac{n(n-1)\dotsm (n-k+1)}{n^k} = 1 \left( 1-\frac{1}{n} \right)\left( 1-\frac{2}{n} \right) \dotsm \left( 1-\frac{k+1}{n} \right), $$ and the latter is a polynomial in $1/n$, with constant term $1$, so it tends to $1$ as required.

$\endgroup$
1
$\begingroup$

Taylor expansion on the RHS

$e^x = 1 + x + \frac 12 x^2 + \frac {1}{3!} x^3 + \cdots$

or

$e^x = \sum_\limits{n=0}^\infty \frac {x^n}{n!}$

binomial expansion on the left-hand side.

$(1+\frac {x}{n})^n =$$ 1 + x + \frac {n-1}{n}x^2 + \frac {(n-1)(n-2)}{n^2} x^3 + \cdots + \frac {(n-1)!}{n^{n-1}} x^n\\ 1 + x + (1-\frac {1}{n})x^2 + (1-\frac {1}{n})(1-\frac {2}n) x^3 + (1-\frac{1}{n})\cdots(1-\frac {n-1}n)x^n$

Every coefficient of $x^n$ on the left-hand side is less than or equal to the corresponding coefficient on the right-hand side.

So we can say for any $n$:

$(1+\frac {x}{n})^n \le e^x$

Choose $m<n$ Consider the first $m$ terms of the expansion of $(1+\frac {x}{n})^n$

Since each term is greater than 0

$ 1 + x + (1-\frac {1}{n})x^2 + (1-\frac {1}{n})(1-\frac {2}n) x^3 + (1-\frac{1}{n})\cdots(1-\frac {m-1}n)x^m \le (1+\frac xn)^n$

Now take the limits as $n\to \infty$ of both sides.

$\sum_\limits{n=1}^m \frac {x^n}{m}\le \lim_\limits{n\to \infty} (1+\frac xn)^n$

So:

$\sum_\limits{n=1}^m \frac {x^n}{n!} \le \lim_\limits{n\to \infty} (1+\frac xn)^n \le \sum_\limits{n=1}^\infty \frac {x^n}{n!}$

Allow $m$ to become large and by the squeeze theorem

$\lim_\limits{n\to \infty} (1+\frac xn)^n = e^x$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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