TITLE: Calculate $\int_0^1 e^x dx$ as a limit of a sum?
QUESTION [8 upvotes]: 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:
$$\int_0^1 e^x\,dx = \frac{e-1}{\lim_{n\to\infty}n\cdot (e^\frac{1}{n} - 1)} = \frac{e-1}{1} = e - 1$$

REPLY [3 votes]: 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.