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