Trying to find the infinite sum of e^-x using integration

[Amad27]

Hello,

I am well aware of the ratio method, and the sum = 1/(1-r) but I want to try this method.

I am trying to understand this:

[itex]\displaystyle \sum_{n=1}^{\infty} e^{-n}[/itex] using integrals, what I have though:

[itex]= \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}[/itex]

So, suppose this is an right-hand Riemann sum, with $m$ *Equal* subintervals.

[itex]f(x_i) = me^{-n}[/itex] represents the *height* of the function, we will have the integral for.

[itex]\Delta(x) = \frac{1}{m}[/itex]

But, How can this be represented as an integral?

Thanks!

 

[quantumtimeleap]

Hi! I'm new here at PF ^^

I believe you're not going to be able to use the Riemann integral definition with a summation of that form. You are trying to use 1/m as the ##\Delta x ## in the definition of the Riemann sum:

## \sum_{n=0}^\infty f(x) \Delta x ##

However 1/m cannot be a difference of two coordinates ## x_{n} - x_{n-1} ##