# Summation as Integral

1. Dec 22, 2008

### f(x)

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 $$\lim_{n \rightarrow \infty } \frac{1}{n}\Sigma_{r=A(x)}^{B(x)}f\frac{r}{n}$$ 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. $$t_r=\frac{r}{1-3r^2+r^4}, \Sigma_{r=1}^{n} t_r = ?$$
Could someone plz explain this method or point me to some resources regarding this.
Thanks

PS: Convergence/Divergence isnt a part of my syllabus, yet.

2. Dec 22, 2008

### sutupidmath

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?

$$\lim_{n\rightarrow \infty} \sum_{r=1}^n\frac{r}{1-3r^2+r^4}$$

Is this correct?

3. Dec 22, 2008

### f(x)

yes the expression is correct, sorry I couldnt do proper LaTEX.
But i am unfamiliar with Riemann Sums, if thats means definite integral, then yes. As I said, I only know the method for f(r/n) , not f(r).
Thanks

4. Dec 22, 2008

### tiny-tim

Hi f(x)!
Surely it's ((B-A)/n)∑r=1n f(A + r(B-A)/n)?

5. Dec 22, 2008

### HallsofIvy

Staff Emeritus
$$\lim_{n\rightarrow \infty} \sum_{r=1}^n\frac{r}{1-3r^2+r^4}$$
is just, by definition,
$$\sum_{r=1}^\infty\frac{r}{1-3r^2+r^4}$$

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

6. Dec 23, 2008

### f(x)

Hello Sir
This is a problem from limits exercises in my textbook. Since Integral isnt possible, am I just supposed to find the sum using algebraic summation techniques and then limit as n-> infinity ?
Could you plz give a hint how to find the sum ?
Thanks

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. Dec 23, 2008

### tiny-tim

Hi f(x)!

You could try partial fractions, and then integrating …

though after that, I get stuck

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

8. Dec 23, 2008

### Dick

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. Dec 23, 2008

### tiny-tim

ah! … got it!

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

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.

Last edited: Dec 24, 2008
10. Dec 23, 2008

### boombaby

hi, could you plz 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

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