# Calculate $\int_0^1 e^x dx$ as a limit of a sum?

As for now, I've been doing the opposite thing. For a given sum in terms of $n\in\mathbb{N}$ I had to calculate the limit (as $n$ approaches infinity) of that sum by applying: http://en.wikibooks.org/wiki/Calculus/Integration_techniques/Infinite_Sums , more precisely, a special case:

$$[a, b] = [0, 1] , \space x_k = \frac{k}{n} \implies \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n f\bigg(\frac{k}{n}\bigg) = \int_0^1f(x)\,dx$$

Example:

$$\lim_{n\to\infty} \bigg( \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}\bigg) = \lim_{n\to\infty} \sum_{k=1}^n\frac{1}{n+k} = \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\frac{1}{1 + \frac{k}{n}}$$

= integral sums (I literally translated the term we use for this in my Analysis class, I don't know how it's called in English) for $f(x) = \frac{1}{1+x}$ on $[0, 1]$

$$= \int_0^1 \frac{1}{1+x} \,dx = \ln(1+x)\bigg\vert_0^1 = \ln2 - \ln1 = \ln2$$

so the limit of the sum as $n$ approaches infinity is $\ln2$.

The following exercise asks to reverse the process (i.e. to use integral sums to calculate the limit), so I have:

$$\int_0^1 e^x \,dx$$

The regular approach gives me:

$\int_0^1 e^x \,dx =$ Leibniz-Newton $= e^x \vert_0^1 = e^1 - e^0 = e -1$

If this integral can be expressed as a limit of a sum, then it should (if I am correct) have the following form:
$$\int_0^1 e^x \,dx = \lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^ne^\frac{k}{n}= \lim_{n \to \infty}\sum_{k=1}^n\frac{e^\frac{k}{n}}{n}$$

And since one definition of $e$ is: $$e = \lim_{n\to\infty}\bigg(1+\frac{1}{n}\bigg)^n$$

I have: $$\int_0^1 e^x \, dx = \lim_{n \to \infty}\sum_{k=1}^n\frac{((1+\frac{1}{n})^n)^\frac{k}{n}}{n} = \lim_{n \to \infty}\sum_{k=1}^n\frac{(1+\frac{1}{n})^k}{n} =$$ $$\lim_{n \to \infty} \bigg[ \frac{(1+\frac{1}{n})^1}{n} + \frac{(1+\frac{1}{n})^2}{n} + \cdots + \frac{(1+\frac{1}{n})^{n-1}}{n} + \frac{(1+\frac{1}{n})^n}{n}\bigg]$$

... which doesn't seem to be what I am looking for (or I am not seeing it).

I would appreciate a hint (or two :))!

EDIT:

By using Ron Gordons big hint we have:

$$\int_0^1 e^x\,dx = \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n e^\frac{k}{n} = \lim_{n\to\infty}\frac{1}{n} \bigg( e^\frac{1}{n} + e^\frac{2}{n} + e^\frac{3}{n} + \cdots + e^\frac{n}{n} \bigg)$$
$$\lim_{n\to\infty}\frac{1}{n} \bigg( e^\frac{1}{n} + e^\frac{1}{n}\cdot e^\frac{1}{n} + e^\frac{1}{n} \cdot e^\frac{1}{n} \cdot e^\frac{1}{n} + \cdots + \underbrace{e^\frac{1}{n} \cdot e^\frac{1}{n} \cdots e^\frac{1}{n}}_{n \space factors} \bigg) =$$ $$\lim_{n\to\infty}\frac{1}{n} \bigg( e^\frac{1}{n} + e^\frac{1}{n} \cdot e^\frac{1}{n} + e^\frac{2}{n} \cdot e^\frac{1}{n} + e^\frac{3}{n} \cdot e^\frac{1}{n} + \cdots + e^\frac{n-1}{n} \cdot e^\frac{1}{n} \bigg) =$$ $$\lim_{n\to\infty}\frac{1}{n} \bigg( e^\frac{1}{n} + \underbrace{(e^\frac{1}{n})^1}_q \cdot e^\frac{1}{n} + \underbrace{(e^\frac{1}{n})^2}_{q^2} \cdot e^\frac{1}{n} + \underbrace{(e^\frac{1}{n})^3}_{q^3} \cdot e^\frac{1}{n} + \cdots + \underbrace{(e^\frac{1}{n})^{n-1}}_{q^{n-1}} \cdot e^\frac{1}{n} \bigg)$$

The expression inside the parentheses can be interpreted as a sum of a geometric series: $a_1 + a_2 + a_3 + a_4 + \cdots + a_n = a_1 + q\cdot a_1 + q^2\cdot a_1 + q^3\cdot a_1 + \cdots + q^{n-1}\cdot a_1$ where
$a_1 = e^\frac{1}{n}$ and $q = e^\frac{1}{n}$ so $a_1 = q$ and by using the formula $S_n = a_1 \cdot \frac{q^n - 1}{q - 1}$ we get:

$$S_n = e^\frac{1}{n} \cdot \frac{(e^\frac{1}{n})^n - 1}{e^\frac{1}{n} - 1} = e^\frac{1}{n} \cdot \frac{e^\frac{n}{n} - 1}{e^\frac{1}{n} - 1} = e^\frac{1}{n} \cdot \frac{e - 1}{e^\frac{1}{n} - 1}$$

so our integral is actually the following limit:

$$\int_0^1 e^x\,dx = \lim_{n\to\infty} \frac{1}{n} \cdot S_n = \lim_{n\to\infty} \frac{1}{n} \cdot e^\frac{1}{n} \frac{e - 1}{e^\frac{1}{n} - 1} =$$ ... by using some properties of limits we get: $$= \lim_{n\to\infty} e^\frac{1}{n} \cdot \lim_{n\to\infty}\frac{1}{n}\frac{e - 1}{e^\frac{1}{n} - 1} = \lim_{n\to\infty} e^\frac{1}{n} \cdot \frac{\lim_{n\to\infty} (e - 1)}{\lim_{n\to\infty} n\cdot (e^\frac{1}{n} - 1)}$$

$$\lim_{n\to\infty} e^\frac{1}{n} = e^\frac{1}{+\infty} = e^0 = 1$$ $$\lim_{n\to\infty} (e - 1) = e - 1$$

so we have:

$$\int_0^1 e^x\,dx = 1\cdot \frac{e - 1}{\lim_{n\to\infty}n\cdot (e^\frac{1}{n} - 1)} = \frac{e - 1}{\lim_{n\to\infty}n\cdot (e^\frac{1}{n} - 1)}$$

The only thing left is to show: $\lim_{n\to\infty}n\cdot (e^\frac{1}{n} - 1) = 1$

Approach 1: $$\lim_{n\to\infty}n\cdot (e^\frac{1}{n} - 1) = \begin{bmatrix} t = \frac{1}{n} \\ n \to \infty \implies t\to 0 \end{bmatrix} = \lim_{t\to 0} \frac{1}{t}(e^t - 1) = \lim_{t\to 0}\frac{e^t - 1}{t} = 1$$

Aproach 2: $$\lim_{n\to\infty}n\cdot (e^\frac{1}{n} - 1) = [\infty \cdot 0] = \lim_{n\to\infty}\frac{(e^\frac{1}{n} - 1)}{\frac{1}{n}} = \bigg[\frac{0}{0}\bigg] = L'Hospital = \lim_{n\to\infty}\frac{(e^\frac{1}{n}\cdot (-\frac{1}{n^2}) - 0)}{-\frac{1}{n^2}} = \lim_{n\to\infty}\frac{e^\frac{1}{n}\cdot (-\frac{1}{n^2})}{-\frac{1}{n^2}} = \lim_{n\to\infty} e^\frac{1}{n} = e^\frac{1}{+\infty} = e^0 = 1$$

finally:

$$\int_0^1 e^x\,dx = \frac{e-1}{\lim_{n\to\infty}n\cdot (e^\frac{1}{n} - 1)} = \frac{e-1}{1} = e - 1$$

## 2 Answers

I think what you want to see is a geometric sum:

$$\sum_{k=0}^n e^{k/n} = \frac{e^{1+1/n}-1}{e^{1/n}-1}$$

Use the fact that

$$\lim_{n\to\infty} n \left (e^{1/n}-1 \right ) = 1$$

and you are almost home.

• Thank you for your answer! I made an edit to my post, could you verify it? – AltairAC May 9 '14 at 15:32
• @Shirohige: You're welcome. Your answer is a little too detailed for me, but it does work. – Ron Gordon May 9 '14 at 15:35
• I see you started with $k=0$ instead of $k=1$. ${}\qquad{}$ – Michael Hardy May 9 '14 at 21:59

You're playing fast and loose with limits when you say that because $\displaystyle\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$, you can conclude that

$$\lim_{n\to\infty}\sum_{k=1}^n\frac{e^{k/n}}{n} = \lim_{n \to \infty}\sum_{k=1}^n\frac{((1+\frac{1}{n})^n)^{k/n}}{n}.$$

In fact $$\int_0^1 e^x \, dx = \lim_{n \to \infty}\sum_{k=1}^n \frac{\left(\lim\limits_{m\to\infty}(1+\frac{1}{m})^m\right)^{k/n}}{n},$$ and because the sum has only finitely many terms, you can write that as $$\int_0^1 e^x \, dx = \lim_{n \to \infty} \lim_{m\to\infty} \sum_{k=1}^n \frac{\left((1+\frac{1}{m})^m\right)^{k/n}}{n}.$$

But I don't think that's promising. As Ron Gordon pointed out in his answer, $\displaystyle\sum_{k=0}^n e^{k/n}$ is a geometric series, so there's a standard formula for its sum.

• Thanks for the explanation but I still have some trouble with that. It seems that I have deceived myself with the term "reverse process" so I literally wanted to do the reverse of what I've been doing in the "sum to integral" exercises. You say in your post "...because the sum has only finitely many terms...". Can we say that because n approaches infinity but never "gets there" or is it because you added another variable "m" so we must fix an arbitrary n (which means that it is finite) and see what happens with the limit as m approaches infinity? – AltairAC May 9 '14 at 15:41
• ... and when I try to evaluate the sum using the limit definition of e (and only one limit, containing n), Wolfram Alpha gives me the correct result: www3.wolframalpha.com/share/… so in terms of intuition and Wolfram Alphas computation result, I don't understand my false conclusion. I've seen a lot of examples where the intuition is misleading and Wolfram Alpha handles things in a wrong manner but this I time, I don't see it. I would appreciate additional explanation on this. – AltairAC May 9 '14 at 16:01
• What if one said that since $\displaystyle\lim_{n\to\infty}\left(1+\frac1n\right)=1$ and $\displaystyle\lim_{n\to\infty}1^n=1$, it follows that $\displaystyle\lim_{n\to\infty}\left(1+\frac1n\right)^n=1$? Clearly that kind of reasoning can yield incorrect results. – Michael Hardy May 9 '14 at 16:25
• I understand. Thank you! – AltairAC May 9 '14 at 17:44