How to Write a Summation as a Riemann-Stieljes Integral

In summary: This is a problem from limits exercises in my textbook. Since Integral isn't possible, am I just supposed to find the sum using algebraic summation techniques and then limit as n-> infinity ?Could you please give a hint how to find the sum ?
  • #1
f(x)
182
0
I have been trying to solve Summation as Limit to Infinity type of questions but there are hardly a few examples I could find in my book
I know the general method for [tex]\lim_{n \rightarrow \infty } \frac{1}{n}\Sigma_{r=A(x)}^{B(x)}f\frac{r}{n}[/tex] where r/n is replaced by x and 1/n by dx, the limits adjusted and integrated.
However, i am unable to understand how to apply this method if the function is f(r) and not of f(r/n)
Eg. [tex] t_r=\frac{r}{1-3r^2+r^4}, \Sigma_{r=1}^{n} t_r = ? [/tex]
Could someone please explain this method or point me to some resources regarding this.
Thanks

PS: Convergence/Divergence isn't a part of my syllabus, yet.
 
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  • #2
HOld on a minute to see whether i am getting u right.

You are saying basically that how would one find the following limit, by recognizing it as a rimann sum right?

[tex]\lim_{n\rightarrow \infty} \sum_{r=1}^n\frac{r}{1-3r^2+r^4}[/tex]

Is this correct?
 
  • #3
sutupidmath said:
HOld on a minute to see whether i am getting u right.

You are saying basically that how would one find the following limit, by recognizing it as a rimann sum right?

[tex]\lim_{n\rightarrow \infty} \sum_{r=1}^n\frac{r}{1-3r^2+r^4}[/tex]

Is this correct?
yes the expression is correct, sorry I couldn't do proper LaTEX.
But i am unfamiliar with Riemann Sums, if that's means definite integral, then yes. As I said, I only know the method for f(r/n) , not f(r).
Thanks
 
  • #4
Hi f(x)! :smile:
f(x) said:
[tex]\lim_{n \rightarrow \infty } \frac{1}{n}\Sigma_{r=A(x)}^{B(x)}f\frac{r}{n}[/tex] where r/n is replaced by x and 1/n by dx, the limits adjusted and integrated.

Surely it's ((B-A)/n)∑r=1n f(A + r(B-A)/n)? :confused:
 
  • #5
[tex]\lim_{n\rightarrow \infty} \sum_{r=1}^n\frac{r}{1-3r^2+r^4}[/tex]
is just, by definition,
[tex]\sum_{r=1}^\infty\frac{r}{1-3r^2+r^4}[/tex]

It can't be written as an integral (unless you use the Riemann-Stieljes integral).
 
  • #6
HallsofIvy said:
[tex]\lim_{n\rightarrow \infty} \sum_{r=1}^n\frac{r}{1-3r^2+r^4}[/tex]
is just, by definition,
[tex]\sum_{r=1}^\infty\frac{r}{1-3r^2+r^4}[/tex]

It can't be written as an integral (unless you use the Riemann-Stieljes integral).
Hello Sir
This is a problem from limits exercises in my textbook. Since Integral isn't possible, am I just supposed to find the sum using algebraic summation techniques and then limit as n-> infinity ?
Could you please give a hint how to find the sum ?
Thanks

tiny-tim said:
Hi f(x)! :smile:


Surely it's ((B-A)/n)∑r=1n f(A + r(B-A)/n)? :confused:
Hello Sir
The standard form I have in my text is the one I have put in my first post, but the one you have posted seems related (i think yours is the one with integral as sum of parts)
 
  • #7
f(x) said:
Hello Sir
This is a problem from limits exercises in my textbook. Since Integral isn't possible, am I just supposed to find the sum using algebraic summation techniques and then limit as n-> infinity ?
Could you please give a hint how to find the sum ?
Thanks
[tex]\sum_{r=1}^\infty\frac{r}{1-3r^2+r^4}[/tex]

Hi f(x)! :smile:

You could try partial fractions, and then integrating …

though after that, I get stuck :redface:

(and don't call us "Sir"!)
 
  • #8
tiny-tim said:
Hi f(x)! :smile:

You could try partial fractions, and then integrating …

though after that, I get stuck :redface:

(and don't call us "Sir"!)

Now why are YOU getting stuck, tiny-tim? That's a great suggestion. Now just complete the squares. One term contains (r+1/2)^2, the other (r-1/2)^2. It telescopes. Doesn't it?
 
  • #9
Dick said:
Now why are YOU getting stuck, tiny-tim? That's a great suggestion. Now just complete the squares. One term contains (r+1/2)^2, the other (r-1/2)^2. It telescopes. Doesn't it?

ah! … got it! :biggrin:

i didn't see that √[(3 + √5)/2] = (1 + √5)/2 :redface:

EDIT: ooh, i didn't need to integrate either …

as soon as you get the right 1/quadratic - 1/quadratic,

you use n2 ± n = n(n ± 1),

and the "telescoping" works immediately. :biggrin:
 
Last edited:
  • #10
HallsofIvy said:
[tex]\lim_{n\rightarrow \infty} \sum_{r=1}^n\frac{r}{1-3r^2+r^4}[/tex]
is just, by definition,
[tex]\sum_{r=1}^\infty\frac{r}{1-3r^2+r^4}[/tex]

It can't be written as an integral (unless you use the Riemann-Stieljes integral).


hi, could you please explain a little about how to write it as an Riemann-Stieljes integral? I learned something about Riemann-Stieljes integral in principle of mathematics but havn't met any concrete examples. Thanks
 

1. What is summation as integral?

Summation as integral is a mathematical concept that relates the process of adding up individual values to the concept of finding the area under a curve. It is often used to approximate the total value of a function or data set.

2. How is summation as integral calculated?

The calculation of summation as integral involves dividing the area under the curve into smaller and smaller rectangles, and then adding up the areas of these rectangles. As the width of the rectangles decreases, the approximation becomes more accurate.

3. What is the difference between summation and integral?

Summation and integral are both mathematical operations that involve adding up values. However, summation is used for discrete data, where the values are distinct and separate, while integral is used for continuous data, where the values are connected and can take on any value within a range.

4. What are the applications of summation as integral?

Summation as integral has many applications in various fields, including physics, engineering, and statistics. It is used to calculate the total value of a function, estimate probabilities, and find the average value of a data set, among others.

5. Are there any limitations to using summation as integral?

While summation as integral is a useful mathematical concept, it is not always an exact calculation and is subject to limitations. It is an approximation and may not give an accurate result if the data is not continuous or if the intervals are not small enough.

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