How to prove the following limit? $$\lim_{n \to \infty} (1+1/n)^n = e$$ I can only observe that the limit should be a very large number!


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    $\begingroup$ Will you please define $e$? $\endgroup$ – Jonas Meyer Jan 5 '12 at 9:50
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    $\begingroup$ @RoupamGhosh: It looks like you are saying that $3$ is the eighth digit after the decimal point. Rather, $e<2.71828183$, but $e=2.718281828459...$. $\endgroup$ – Jonas Meyer Jan 5 '12 at 10:02
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    $\begingroup$ 25 minutes. $ $ $\endgroup$ – Did Jan 5 '12 at 10:22
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    $\begingroup$ @Didier : ????? $\endgroup$ – Patrick Da Silva Jan 5 '12 at 10:32
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    $\begingroup$ @Patrick: this is the time span between the moment when the question was posted and the moment when an answer got accepted. For a discussion, see the comments on the main post there. $\endgroup$ – Did Jan 5 '12 at 12:16

Just apply the binomial theorem then move things around a bit: $$\left(1+\tfrac{1}{n}\right)^n = \sum_{k=0}^n \frac{{n \choose k}}{n^k} = \sum_{k=0}^n \frac{1}{k!} \frac{n!/(n-k)!}{n^k}.$$

Now if you could just get rid of that $\frac{n!/(n-k)!}{n^k}$ term... (using the idea of $n \to \infty$ and $k$ is usually small in comparison)

edit Jonas makes a good point, I was implicity assuming the definition $e = \sum_{k = 0}^\infty \frac{1}{k!}$

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    $\begingroup$ Where is the proof? $\endgroup$ – Did Jan 5 '12 at 10:22
  • $\begingroup$ @DidierPiau, not here. This is just a starting point for the OP. $\endgroup$ – user16697 Jan 5 '12 at 10:48
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    $\begingroup$ You've at least shown that $\left(1+\frac{1}{n}\right)^n\le\sum\limits_{k=0}^\infty\frac{1}{k!}$. $\endgroup$ – robjohn Jan 5 '12 at 11:50

Actually, the way things work out in mathematics usually is that we only prove that $x_n = (1+1/n)^n$ is a convergent sequence, and we define its limit to be $e$. Some people use other definitions for $e$ and show that it is equivalent to this definition, but there are many ways to do this that are logically equivalent.

(Just for the record, you don't "prove" limits, you compute them.)

One way to show that this sequence is convergent is to show that it is increasing and bounded above. This is a bit technical but works out quite well. To prove that $x_n$ is increasing, we show that $x_{n+1} \ge x_n$, which is equivalent to $x_{n+1}/x_n \ge 1$ : \begin{align} \frac{x_{n+1}}{x_n} & = \frac{ \left( 1 + \frac 1{n+1} \right)^{n+1} }{\left( 1 + \frac 1n \right)^n } = \left( 1 + \frac 1{n+1} \right) \left( \frac{ 1 + \frac 1{n+1} }{1 + \frac 1n} \right)^n = \left( \frac {n+2}{n+1} \right) \left( \frac{(n+2)n}{(n+1)^2} \right)^n \\ & = \left( \frac{n+2}{n+1} \right) \left( 1 + \frac{(n+2)n - (n+1)^2 }{(n+1)^2} \right)^n \\ & = \left( \frac{n+2}{n+1} \right) \left( 1 - \frac{1}{(n+1)^2} \right)^n \\ & \ge \left( \frac{n+2}{n+1} \right) \left( 1 - \frac n{(n+1)^2} \right) \\ & \ge \left( \frac{n+2}{n+1} \right) \left( \frac{n^2 + n + 1}{n^2 + 2n + 1} \right) = \frac{n^3 + 3n^2 + 3n + 2}{n^3 + 3n^2 + 3n + 1} \ge 1. \end{align}

The first inequality holds by Bernoulli's inequality ( I'm referring to $(1+x)^n \ge 1+nx$ for $x \ge -1$).

To prove that $x_n$ is bounded above, we use the binomial theorem :

\begin{align} \left(1+\frac 1n\right)^n & = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} \frac 1{n^k} = \sum_{k=0}^n \frac 1{k!} \frac{n!}{n^k(n-k)!} \\ & \le 1 + \sum_{k=1}^n \frac 1{k!} \le 1 + \sum_{k=1}^n \frac 1{2^{k-1}} = 1 + \frac{1- \frac 1{2^n}}{ 1- \frac 12} < 1 + 2 = 3. \\ \end{align} Note that I used the fact that $k! \ge 2^{k-1}$, which implies $\frac 1{2^{k-1}} \ge \frac 1{k!}$. (To see this, just write $k! = k \times \dots \times 2 \times 1 \ge 2 \times \dots \times 2 \times 1 = 2^{k-1}$.)

Another way would be to show that this sequence converges to $\sum_{k=0}^{\infty} \frac 1{k!}$ (which is another definition for $e$) by expanding $(1+1/n)^n$ with the binomial theorem.

Hope that helps,

  • $\begingroup$ Here, the OP needs to prove that the equality holds (IMO). +1, anyway. $\endgroup$ – Gigili Jan 5 '12 at 10:23
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    $\begingroup$ I've explained that he can only prove that equality holds if he chooses a definition for $e$. I can't answer completely since he didn't choose one. $\endgroup$ – Patrick Da Silva Jan 5 '12 at 10:31
  • $\begingroup$ Right, I missed that part. The parentheses distracted my attention. $\endgroup$ – Gigili Jan 5 '12 at 10:36
  • $\begingroup$ Since it seems that the OP thinks that $\left(1+\frac{1}{n}\right)^n$ gets very large, I agree that they need to see a proof that the limit exists. (+1) $\endgroup$ – robjohn Jan 5 '12 at 11:57
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    $\begingroup$ Note that at some point in my upperbounding of the sequence, the expression for the partial sum of the value of $e$ appears. I could've just bounded above by the series (which is $e$) but that would become (a little) circular (not technically, but morally). $\endgroup$ – Patrick Da Silva Jan 6 '12 at 18:05

The fact that it converges to a limit is already explained in other answers. Then to get an idea as to why this limit must match the definition of $e$:

We know that $\exp(x) = e^x$ is a function that starts with $\exp(0) = 1$, and grows with a rate that is equal to it's current value (ie, $\frac{d}{dx}e^x = e^x$). So a good way to find this, is to approximate it with a finite number of segments, where the rate at each segment is constant, and equal to the value at the beginning of the segment:

$\exp_n(0) = 1$

$\exp_n(t) = \exp_n(t - \frac{1}{n}) + \frac{1}{n}\exp_n(t - \frac{1}{n})$

$\exp_n(t) = (1 + \frac{1}{n})\exp_n(t - \frac{1}{n})$

Clearly, taking smaller and smaller segments will give a better approximation of the actual smooth function, with a constantly changing rate, and in the limit, it will match the definition of $\exp$. Unrolling the recursion for $\exp_n(1)$ clearly gives

$\exp_n(1) = (1 + \frac{1}{n})^n \exp_n(0) = (1 + \frac{1}{n})^n$


$\exp(1) = \lim_{n \rightarrow \infty} \exp_n(1) = \lim_{n \rightarrow \infty} (1 + \frac{1}{n})^n$

This is enough to prove that if the limit exists, then it matches the definition of $e$.

  • $\begingroup$ Watch out since "unrolling the recursion of $\mathrm{exp}$" doesn't give you the equality in your last line, but an approximation. Great answer though. +1 $\endgroup$ – Patrick Da Silva Jan 5 '12 at 10:22
  • $\begingroup$ How you got the result that $\exp(t) = \exp(t - \frac{1}{n}) + \frac{1}{n}\exp(t - \frac{1}{n})$? $\endgroup$ – Gigili Jan 5 '12 at 10:27
  • $\begingroup$ This is of course not true for the actual $\exp$ function, but only in our approximations. Basically, it says that the value at the end of a linear segment (of length $\frac{1}{n}$) equals the value at the beginning of the segment, plus its length times its rate of change (which, by definition, is equal to the value at the beginning of the segment). $\endgroup$ – Lieven Jan 5 '12 at 10:42
  • $\begingroup$ @Lieven: Aha, got it. Thank you. $\endgroup$ – Gigili Jan 5 '12 at 14:02

One way to prove it is to manipulate the the exponential of the logarithm. Note that we can write: \begin{equation*} \lim_{n\to\infty} e^{n\ln(1+\frac{1}{n})}=e^{\lim_{n\to\infty}n\ln (1+\frac{1}{n})}=e^{\lim_{n\to\infty}\frac{\ln(1+\frac{1}{n})}{1/n}}. \end{equation*} An application of L'Hopital's rule gives \begin{equation*} e^{\lim_{n\to\infty}\frac{1}{1+1/n}}=e^1=e.~_{\square} \end{equation*}


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