- #1
Amad27
- 412
- 1
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!
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!
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