Computing a Series Sum: n2 + (r-1)2 to n2 + (n-1)2

Click For Summary
The discussion focuses on computing the sum of a specific series involving terms of the form 1/(n² + (r-1)²) to 1/(n² + (n-1)²). Participants explore whether a closed-form expression or an approximation is needed and discuss the relevance of Riemann sums and definite integrals to the problem. They highlight the differences between two related problems, noting that the definitions of αn and βn differ in their summation approaches. The conversation emphasizes the importance of understanding the relationship between integration and summation, as well as the need for clarity in problem statements to avoid confusion. Overall, the complexity of the series and its connection to calculus concepts is underscored.
andyrk
Messages
658
Reaction score
5
Can someone explain how to compute the sum of the following series?
1/(n2 + (r-1)2) + 1/(n2 + (r)2) + 1/(n2 + (r+1)2) + ...1/(n2 + (n-1)2)
 
Mathematics news on Phys.org
Is anyone there?
 
Explain the context. Do you seek a closed form expression that is simpler to write than the sum itself? Do you need an approximation?

I suggest we first try to find the summation::
S_m = \sum_{k=1}^m {\frac{1}{1 + k^2}}.
since the result might be used to derive the summation of the series you gave.
 
I don't have an idea as to how to go about finding even the summation you gave. Can you give me a clue?
 
I can only give suggestions. ( I could give a "clue" if I knew how to sum it already.)

Apply the section "Rational Functions" in the article: http://en.wikipedia.org/wiki/List_of_mathematical_series

\sum_{k=1}^\infty \frac{1}{1+k^2} = -\frac{1}{2} + \frac{\pi}{2} coth(\pi)

So perhaps a substitution involving the hyperbolic cotangent would be useful in doing the finite sum. I'll keep thinking about it.
 
Woah. I think this is beyond what my curriculum asks for. I am pretty sure that it doesn't involve trigonometric solutions as simplification of any series. I think this has something else to do with. Definitely not this way.
 
What has your curriculum covered? Is this in a chapter on mathematical induction?
 
Stephen Tashi said:
What has your curriculum covered? Is this in a chapter on mathematical induction?
Mathematical Induction in my curriculum doesn't include sum of series etc. It just involves proving LHS = RHS or proving the given statement by using induction. This topic is done in sequences and series. But it just involves basic sums like - ∑n, ∑n2, ∑n3. And I don't know how to incorporate these 3 into the summation above to simplify it.
 
Did you quote the problem exactly? - or does the statement of the problem use summation notation?
 
  • #10
Stephen Tashi said:
Did you quote the problem exactly? - or does the statement of the problem use summation notation?
Its a part of the problem that I told you. The exact problem is-

Consider a function f(x) = 1/(1+x2)
Let αn = 1/n *(\sum_{r=1}^n {f(\frac{r}{n}}))
and βn = 1/n *(\sum_{r=0}^{n-1} {f(\frac{r}{n}})) ; n ∈ N
α = limn →∞n ) & β = limn →∞n )
then αn/β - βn/α will always be? (Answer: A real number)
 
Last edited:
  • #11
andyrk said:
and βn = 1/n *(\sum_{r=0}^n-1 {f(\frac{r}{n}}) ; n ∈ N

Is the sum supposed to be \sum_{r=0}^{n-1} f(\frac{r}{n}) ? It begins at r = 0 ?
 
  • #12
Stephen Tashi said:
Is the sum supposed to be \sum_{r=0}^{n-1} f(\frac{r}{n}) ? It begins at r = 0 ?
Yes. I didn't know how to write it up. Anyways, I have edited it now. :)
 
  • #13
My guess is that you don't have to find a summation formula in order to work the problem. I'd start wtih an equation relating \alpha_n to \beta_n

\alpha_n + \frac{1}{n} f(\frac{0}{n}) - \frac{1}{n}f(\frac{n}{n}) = \beta_n
 
  • #14
Stephen Tashi said:
My guess is that you don't have to find a summation formula in order to work the problem. I'd start wtih an equation relating \alpha_n to \beta_n

\alpha_n + \frac{1}{n} f(\frac{0}{n}) - \frac{1}{n}f(\frac{n}{n}) = \beta_n

How did you come about this equation? I can't figure it out?
 
  • #15
Compare the terms in the sum for \alpha_n with those in the sum for \beta_n. The sum for \alpha_n is missing the r= 0 term that \beta_n has. The sum for \alpha_n has a term for r = n that \beta_n does not.
 
  • #16
Oh yeah. I got that now. But the solution that I have says-
\alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n))
And comparing to the clue you gave me I get-
\alpha_n - \beta_n = \frac{1}{n}( f(1) - f(0)
This leaves me confused as to what went wrong.
Here's the solution that I have. It confuses me everytime I go through it.
 

Attachments

  • Solution.jpg
    Solution.jpg
    25.4 KB · Views: 451
Last edited:
  • #17
andyrk said:
Can someone explain how to compute the sum of the following series?
1/(n2 + (r-1)2) + 1/(n2 + (r)2) + 1/(n2 + (r+1)2) + ...1/(n2 + (n-1)2)
No:D
 
  • #18
andyrk said:
And comparing to the clue you gave me I get-
\alpha_n - \beta_n = \frac{1}{n}( f(1) - f(0)

There are two different problems. The problem you stated is related to Riemann sums for \int_0^1 f(x) dx. The problem in the solution is related to Riemann sums for \int_0^n f(x) dx.

In the problem you stated, the definition of \alpha_n is a Riemann sum using the altitude at the right hand point of the base of each rectangle. In the solution, \alpha_n is defined to use the altitutde at the left hand point of the base of each rectangle.
 
  • #19
Woah. This is not even a part of my course. I think you went a bit too far with that. Could you please explain what you said more clearly? How does the limit n→∞ change the limits from 0 to n to 0 to 1 in the integral?
 
  • #20
Have you studied definite integrals? Riemann sums?
 
  • #21
I have studied definite integrals. Is Riemann sums just another name of it?
 
  • #22
Definite integrals are usually defined as the limit of an approximation process that divides he area under the graph of a function into rectangles and sums their areas. Such sums are the Riemann sums.
 
  • #23
Yup. I know that alright. But still I am not able to understand how the limits change from (0 to n) to (0 to 1)?
 
  • #24
How they change between the two problems is just arbitrary notation. Why they change between \alpha_n and \beta_n is because the symbols \alpha_n, \beta_n are used to denote two different Riemann sums.

One sum computes the area of a rectangle with a given base by using the value of the function at the left endpoint of the base. The other sum computes the area of a rectangle with a given base by using the value of the function at the right endpoint. In the particular problem we have an integral from x = 0 to x = 1. The sum that begins with the term \frac{1}{n} f(\frac{0}{n}) is using the left endpoint of the base as the height of the rectangle. That sum ends with the term \frac{1}{n} f(\frac{n-1}{n}) because the last rectangle in the interval [0,1] has base [\frac{n-1}{n}, \frac{n}{n} ]. So when we we define limits for the summation using an index k , the index goes from k = 0 to k = n-1.

The other Riemann sum computes the are of a given base by using the value of the function at the right end of the base. You should be able to see why an index k for that sum would go from k = 1 to k = n.
 
  • #25
andyrk said:
I am not able to understand how the limits change from (0 to n) to (0 to 1)?

If you mean the limits of integration, they don't change for x. I was incorrect to say that. The solution page you has introduces a third variable "r" that is also marked on the x-axis and their picture seems to indicated that r goes from zero to infinity.
 
  • #26
My questions are:-
(1) Why is α = limn→∞n) = \int_0^1 f(x) dx ? Is this related to the graphs provided in the solution including the area under the curve using rectangular strips?
(2) Either \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n)) or \alpha_n - \beta_n = \frac{1}{n}( f(1) - f(0)
How can both be true? And if anyone of them is correct then why is the other one incorrect?
 
  • #27
andyrk said:
My questions are:-
(1) Why is α = limn→∞n) = \int_0^1 f(x) dx ? Is this related to the graphs provided in the solution including the area under the curve using rectangular strips?

The definition of a definite integral is the limit of the Riemann sums as the number of rectangles approaches infinity (which makes the dimensions of their bases approach zero). A typical calculus book has a section about this, sometimes an entire chapter.

(2) Either \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n)) or \alpha_n - \beta_n = \frac{1}{n}( f(1) - f(0)
How can both be true? And if anyone of them is correct then why is the other one incorrectt

They apply to different problems. Compare the notation you used in your post #10 with the notation used in the image you attached in post #16. The symbol \alpha_n is defined differently in post #10 than in post #16. The symbol \beta_n is also defined differently.
 
  • #28
Stephen Tashi said:
The definition of a definite integral is the limit of the Riemann sums as the number of rectangles approaches infinity (which makes the dimensions of their bases approach zero). A typical calculus book has a section about this, sometimes an entire chapter.

They apply to different problems. Compare the notation you used in your post #10 with the notation used in the image you attached in post #16. The symbol \alpha_n is defined differently in post #10 than in post #16. The symbol \beta_n is also defined differently.

That is really strange. Because both the question as well as the solution belong to the same source from my course. So ideally, the solution should represent and should use the same information that is provided in the question. But as you are saying that the solution and the question give different information about the variables involved, is it safe for me to come to the conclusion that the question is wrong and can be left out?
 
  • #29
andyrk said:
is it safe for me to come to the conclusion that the question is wrong and can be left out?

One could equally well say the question is right and the solution is wrong. Whether you can leave it out is matter of academic policy, your instructors wishes, customary procedues etc. These aren't things I know about.

I think a capable student could learn enough from the example solved by the solution page to solve a different but similar question.
 
  • #30
Stephen Tashi said:
I think a capable student could learn enough from the example solved by the solution page to solve a different but similar question.
But how do I know whether the solution is right or not? I could learn from the solution only if it is consistent with the problem/question. Otherwise, I am not able to make anything out of the solution at all. Because it simply doesn't refer to the question that is given.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K