Finding the infinite sum of $e^{-n}$ using integrals

Question

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!

Answer 1

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 $$\sum_{n=1}^{+\infty}e^{-n} = \dfrac{e^{-1}}{1-e^{-1}} = \frac{1}{e-1}$$

Answer 2

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 $$\sum_{n=1}^\infty e^{-n}={e\over e-1}\int_1^\infty e^{-x}\>dx={e\over e-1}\cdot {1\over e}={1\over e-1}\ .$$