MHB Finding an Integral for a given Riemann Sum

[Amad27]

Hello,

This question is purely inspired by: http://mathhelpboards.com/calculus-10/evaluating-infinite-sum-e-x-using-integrals-12838.html

My other question. Anyhow,

How do you find the integral for a given specific Riemann sum.

Suppose the same one given in the link;

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

How can someone convert that into an integral?

We know $\Delta(x) = \frac{1}{m}$. So

$me^{-x}$, is the height of the function in some subinterval.

This is possibly a right hand Riemann sum.

IdeaS? Thanks!

 

[Euge]

Hello,

This question is purely inspired by: http://mathhelpboards.com/calculus-10/evaluating-infinite-sum-e-x-using-integrals-12838.html

My other question. Anyhow,

How do you find the integral for a given specific Riemann sum.

Suppose the same one given in the link;

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

How can someone convert that into an integral?

We know $\Delta(x) = \frac{1}{m}$. So

$me^{-x}$, is the height of the function in some subinterval.

This is possibly a right hand Riemann sum.

IdeaS? Thanks!

No, it cannot be done with this method. The limit parameter $m$ is independent of of the summation index $x$, so $\frac{1}{m}\sum_{x = 1}^m me^{-x}$ is not a Riemann sum.

 

[Amad27]

No, it cannot be done with this method. The limit parameter $m$ is independent of of the summation index $x$, so $\frac{1}{m}\sum_{x = 1}^m me^{-x}$ is not a Riemann sum.

Can this sum then be done with integration or differentiation at all?

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

WITHOUT using the $S = \frac{1}{1-r}$ method?

 

[Euge]

Can this sum then be done with integration or differentiation at all?

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

WITHOUT using the $S = \frac{1}{1-r}$ method?

You can represent the series as an integral, but it won't be useful for evaluation. Geometric series are used as a basis for many differentiation and integration methods, so you'll have problems trying to do the reverse.

 

[Amad27]

You can represent the series as an integral, but it won't be useful for evaluation. Geometric series are used as a basis for many differentiation and integration methods, so you'll have problems trying to do the reverse.

@Euge, that is fairly interesting. So, how can I write the summation as an integral? I just want to see! Thanks!

 

[Euge]

@Euge, that is fairly interesting. So, how can I write the summation as an integral? I just want to see! Thanks!

Let

$$u(x) = \begin{cases}0, & x \le 0\\1, & x > 0\end{cases}.$$

Set

$$\alpha(x) = \sum_{n = 1}^\infty e^{-n}u\Bigl(x - \frac{1}{n}\Bigr).$$

Then

$$\sum_{n = 1}^\infty e^{-n} = \int_0^1 1\, d\alpha.$$

This is a representation of your series as a Riemann-Stieltjes integral. More generally, if $f$ is continuous on $[0,1]$, then

$$\sum_{n = 1}^\infty e^{-n} f\bigl(\tfrac1{n}\bigr) = \int_0^1 f\, d\alpha.$$

 

[Amad27]

Let

$$u(x) = \begin{cases}0, & x \le 0\\1, & x > 0\end{cases}.$$

Set

$$\alpha(t) = \sum_{n = 1}^\infty e^{-n}u\Bigl(x - \frac{1}{n}\Bigr).$$

Then

$$\sum_{n = 1}^\infty e^{-n} = \int_0^1 1\, d\alpha.$$

This is a representation of your series as a Riemann-Stieltjes integral. More generally, if $f$ is continuous on $[0,1]$, then

$$\sum_{n = 1}^\infty e^{-n} f\bigl(\tfrac1{n}\bigr) = \int_0^1 f\, d\alpha.$$

How? The sum does not equal $1$ by any means. In general,

Is there any method using perhaps integrals to derive that sum?

Like use integrals to get to a sum, perhaps "solve" for the sum? Using integration to find a definite integral related to the sum??

Thanks!

 

[Euge]

How? The sum does not equal $1$ by any means.

Like I said, the integral representing your series is a Riemann-Stieltjes integral. It's a generalization of the Riemann integral. When $\alpha(x) = x$, a Riemann-Stieltjes integral $\int_a^b f\, d\alpha$ reduces to the ordinary Riemann integral $\int_a^b f(x)\, dx$.

 

[Amad27]

Like I said, the integral representing your series is a Riemann-Stieltjes integral. It's a generalization of the Riemann integral. When $\alpha(x) = x$, a Riemann-Stieltjes integral $\int_a^b f\, d\alpha$ reduces to the ordinary Riemann integral $\int_a^b f(x)\, dx$.

Hi,

Thanks a lot. That's good to know. I made a new thread called: Using integral methods to evaluate any summation. Perhaps you'll be interested? Thanks @Euge!