Evaluating infinite sum for e^(-x) using integrals

[Amad27]

Hello,

I have began my journey on infinite sums, which are very interesting. Here is the issue:

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!

 

[alyafey22]

I am not sure this can be solved using integration. The standard way to solve it using Geometric series

$$\sum_{n\geq 0} x^n = \frac{1}{1-x} \,\,\,\, |x| < 1$$

The proof is not difficult if you are interested.

 

[Amad27]

I am not sure this can be solved using integration. The standard way to solve it using Geometric series

$$\sum_{n\geq 0} x^n = \frac{1}{1-x} \,\,\,\, |x| < 1$$

The proof is not difficult if you are interested.

Really? I was really hoping it was. It there a way to write the Riemann Sum:

$\displaystyle \lim_{m\to\infty}\frac{1}{m}\sum_{n=1}^{m} me^{-n}$

As an integral form?

 

[Amad27]

Really? I was really hoping it was. It there a way to write the Riemann Sum:

$\displaystyle \lim_{m\to\infty}\frac{1}{m}\sum_{n=1}^{m} me^{-n}$

As an integral form?

I thought on this on my own.

What integral evaluate to $e^{-n}$?

$\displaystyle e^{-n} = \int_{0}^{e} -nx^{-(n+1)} \,dx$

Originally, we had, $\displaystyle \sum_{n=1}^{\infty} e^{-n}$ This is:

$\displaystyle \sum_{n=1}^{\infty}\int_{0}^{e} -nx^{-(n+1)} \,dx$$\displaystyle e^{-n} = \int_{0}^{e} -nx^{-(n+1)} \,dx$

Originally, we had, $\displaystyle \sum_{n=1}^{\infty} e^{-n}$ This is:

$\displaystyle \int_{0}^{e} \sum_{n=1}^{\infty} -nx^{-(n+1)} \,dx$

$S_1 = -x^{-2}$
$S_2 = -x^{-2} - 2x^{-3}$
$S_3 = -x^{-2} - 2x^{-3} - 3x^{-4}$
$S_4 = -x^{-2} - 2x^{-3} - 3x^{-4} - 4x^{-5}$

If someone can help me find the sum, I can finish this problem.

 

[skeeter]

To reiterate that posted by ZaidAlyafey ...

The sum of an infinite geometric series is $$\frac{a}{1-r}$$ , where $$a$$ is the first term and $$r$$ is the common ratio such that $$|r| < 1$$

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