Finding the infinite sum of $e^{-n}$ using integrals I am trying to understand this: 
$\displaystyle \sum_{n=1}^{\infty} e^{-n}$ using integrals, what I have though:
$= \displaystyle \lim_{m\to\infty} \sum_{n=1}^{m} e^{-n}$
$= \displaystyle \lim_{m\to\infty} \frac{1}{m}\sum_{n=1}^{m} me^{-n}$
So, suppose this is an right-hand Riemann sum, with $m$ Equal subintervals. 
$f(x_i) = me^{-n}$ represents the height of the function, we will have the integral for. 
$\Delta(x) = \frac{1}{m}$
But, How can this be represented as an integral? 
Thanks!
 A: Maybe it's not what you want, but it uses integral method...
We compute the integral $$\int_0^{e^{-1}}1 dx$$ using the partition $\{e^{-n}, n\geq 1\}$, since the integrand is constant we have 
$$\int_0^{e^{-1}}1 dx = \sum_{n=1}^{+\infty}\left(e^{-n} - e^{-(n+1)}\right)$$
i.e. $$e^{-1} = \sum_{n=1}^{+\infty}\left(e^{-n} - e^{-(n+1)}\right) = (1-e^{-1})\sum_{n=1}^{+\infty}e^{-n}$$
so $$\sum_{n=1}^{+\infty}e^{-n} = \dfrac{e^{-1}}{1-e^{-1}} = \frac{1}{e-1}$$
A: In the following I shall not convert the required sum into a Riemann sum, but use the  primitive $x\mapsto -e^{-x}$ of $x\mapsto e^{-x}$ to sum the series without "algebraic tricks". From
$$\int_n^{n+1}e^{-x}\>dx=-e^{-x}\biggr|_n^{n+1}=e^{-n}\left(1-{1\over e}\right)\qquad(n\geq1)$$
it follows that
$$\int_1^\infty e^{-x}\>dx=\sum_{n=1}^\infty\int_n^{n+1}e^{-x}\>dx=\left(1-{1\over e}\right)\sum_{n=1}^\infty e^{-n}\ .$$
Solving for the sum in question we obtain
$$\sum_{n=1}^\infty e^{-n}={e\over e-1}\int_1^\infty e^{-x}\>dx={e\over e-1}\cdot {1\over e}={1\over e-1}\ .$$
