Ok so we know that:

e= lim n→∞(1+1/n)^n.

and we know by binomial theorem, that

lim n→∞ of $\sum_{k=0}^n {n \choose k} (1/n)^k = (1+ \frac{1}{n})$

To simplify further to $\sum_{k=0}^n \frac{1}{k!} = e$

we must evaluate the following limit:

n→∞${n \choose k} \frac{1}{n^k}=\frac{(n)(n-1)(n-2)...(n-k+1)}{k! n^k}$

and this is supposed to equal to $\frac{1}{k!}$

This is not clear to me algebriacally, so may some one please clear this up step by step so I can understand?


We have $$e=\lim_{n\to\infty}\sum_{k=0}^n\frac{\binom{n}{k}}{n^k}=\lim_{n\to\infty}\sum_{k=0}^\infty\frac{\binom{n}{k}}{n^k}=\lim_{n\to\infty}\sum_{k=0}^\infty\frac{1}{k!},$$where with $k$ fixed$$\lim_{n\to\infty}\frac{\binom{n}{k}}{n^k}=\frac{1}{k!}\lim_{n\to\infty}\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)=\frac{1^{k-1}}{k!}=\frac{1}{k!}.$$

  • $\begingroup$ Yes but why is (n choose k)/n^k = 1/k! ? $\endgroup$ – GLaDOS Jan 27 at 16:06
  • $\begingroup$ @GLaDOS It's not; it's $k!^{-1}\prod_{1\le j\le k-1}(1-j/n)$. $\endgroup$ – J.G. Jan 27 at 16:07
  • $\begingroup$ Why is that product equal to n!/n^k(n-k)! then? $\endgroup$ – GLaDOS Jan 27 at 16:10
  • $\begingroup$ @GLaDOS Because $n!/(n-k)!=n\cdots(n-k+1)$. $\endgroup$ – J.G. Jan 27 at 16:11
  • $\begingroup$ What I am unclear about is why: n^k(n)(n-1)(n-2)...(n-k+1) = (1-1/n)(1-2/n)(1-3/n)...(1-(k-1)/n) $\endgroup$ – GLaDOS Jan 27 at 16:20

We may prove that by considering the Maclaurin series of the exponential function.$$f(x)=e^x=\displaystyle \sum_{k=0}^{\infty}\dfrac{x^k}{k!}$$ $$\therefore f(1)=e=\displaystyle \sum_{k=0}^{\infty}\dfrac{1}{k!}=1+\dfrac{1}{2!}+\dfrac{1}{3!}+\cdots$$

  • $\begingroup$ Prove the maclaurin series $\endgroup$ – GLaDOS Jan 27 at 16:06
  • $\begingroup$ Maclaurin series is the Taylor series centered at $x=0$. Maclaurin series of a function $f(x)$ is defined as $\displaystyle \sum_{k=0}^{\infty} a_k x^k$ where $a_k=\frac{f^{(k)}(0)}{k!}$. You can figure it out for $e^x$ which is an extremely easy exercise. $\endgroup$ – Paras Khosla Jan 27 at 16:10
  • $\begingroup$ Prove the taylor series $\endgroup$ – GLaDOS Jan 27 at 16:10
  • $\begingroup$ You may refer to this link: en.wikipedia.org/wiki/Taylor_series $\endgroup$ – Paras Khosla Jan 27 at 16:11
  • $\begingroup$ It only includes a definition, but ok. I guess I can't ask much from you then. $\endgroup$ – GLaDOS Jan 27 at 16:13

Since you know that

$$\lim_{n\rightarrow \infty}\frac{c_0 + c_1n + \cdots + c_kn^k}{n^k} = c_k$$

for any constants $c_i$, we get

$$\lim_{n\rightarrow \infty}\frac{c_0 + c_1n + \cdots + c_kn^k}{k!n^k} = \frac{1}{k!}\lim_{n\rightarrow \infty}\frac{c_0 + c_1n + \cdots + c_kn^k}{n^k} = \frac{c_k}{k!}$$

Note that your numerator, when expanded, is a monic polynomial with degree $k$, which fits the above with $c_k = 1$.


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