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

[andyrk]

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)

 

[andyrk]

Is anyone there?

 

[Stephen Tashi]

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.

 

[andyrk]

I don't have an idea as to how to go about finding even the summation you gave. Can you give me a clue?

 

[Stephen Tashi]

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.

 

[andyrk]

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.

 

[Stephen Tashi]

What has your curriculum covered? Is this in a chapter on mathematical induction?

 

[andyrk]

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.

 

[Stephen Tashi]

Did you quote the problem exactly? - or does the statement of the problem use summation notation?

 

[andyrk]

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 α_n = 1/n *(\sum_{r=1}^n {f(\frac{r}{n}}))
and β_n = 1/n *(\sum_{r=0}^{n-1} {f(\frac{r}{n}})) ; n ∈ N
α = lim_{n →∞}(α_n ) & β = lim_{n →∞}(β_n )
then α_n/β - β_n/α will always be? (Answer: A real number)

 

[Stephen Tashi]

and β_n = 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 ?

 

[andyrk]

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. :)

 

[Stephen Tashi]

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

 

[andyrk]

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?

 

[Stephen Tashi]

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.

 

[andyrk]

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.

 

[zoki85]

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

 

[Stephen Tashi]

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.

 

[andyrk]

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?

 

[Stephen Tashi]

Have you studied definite integrals? Riemann sums?

 

[andyrk]

I have studied definite integrals. Is Riemann sums just another name of it?

 

[Stephen Tashi]

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.

 

[andyrk]

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)?

 

[Stephen Tashi]

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.

 

[Stephen Tashi]

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.

 

[andyrk]

My questions are:-
(1) Why is α = lim_n→∞(α_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?

 

[Stephen Tashi]

My questions are:-
(1) Why is α = lim_n→∞(α_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.

 

[andyrk]

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?

 

[Stephen Tashi]

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.

 

[andyrk]

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.

 

[Stephen Tashi]

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.

The solution deals with a problem nearly identical to the one that is given. If you comprehend the subject matter, you should be able to work problems after seeing similar but not identical problems solved. I don't know how your course is taught. Perhaps you are being taught to do rote memorization of solutions to a certain list of problems and have not had the opportunity to develop independent judgement.

From the perspective of a typical USA introductory calculus class, the problem is hard problem because it requires understanding the relation between integration and Riemann sums, understanding summation notation, and attention to detail. I think it would take an "A" student to understand the solution page. However, after the "A" student read the solution page, I'd expect the student to be able to work the problem that was given.

 

[andyrk]

Perhaps you are being taught to do rote memorization of solutions to a certain list of problems and have not had the opportunity to develop independent judgement.

I can assure you that I am definitely not being taught to rote memorize things. This is just one of the several such (and even more difficult) questions given in my course. And I agree, that the problem is similar but still doesn't make sense. It is similar because it is using the same function f(x), same variables α_n, β_n, α, β. But after that it is pretty confusing. This is because, the question doesn't involve any integral with limits 0 to 1. It involves a summation (which is very much different from an integral) with limits 0 to n-1 and 1 to n. I don't understand how could you compare these differences and call them similar or identical?

 

[Stephen Tashi]

What is the Riemann sum \alpha_n for approximating \int_a^b f(x) d(x) that uses n rectangles with bases of equal length and uses the value of f(x) at the right hand endpoint for the altitude of the rectangle?

The length of each base is \triangle x = \frac{(b-a)}{n}

\alpha_n = \sum_{k=1}^n \triangle x \ f( a + k \triangle x) = \triangle x \sum_ {k=1}^n f(a + k \triangle x)

This sum is associated with a picture showing the graph of a function and the rectangles that are involved in the sum.

The Riemann sum \beta_n that uses the same bases but takes the altitudes to be the values of the function at the left hand endpoints of the bases is:

\beta_n = \sum_{k=0}^{n-1} \triangle x\ f(z + k \triangle x) = \triangle x \sum _{k=0}^{n-1} f(a + k \triangle x)

The definition of \int_a^b { f(x) dx} says (among other things) that it is the limit of either type of Riemann sum as n approaches infinity.

Let \alpha = \lim_{n\rightarrow \infty} \alpha_n then \alpha = \int_a^b f(x) dx
Let \beta = \lim_{n\rightarrow \infty} \beta_n then \beta = \int_a^b f(x) dx

That is relation between the sums and the integral.

Your problem is a particular case with:
a = 0
b = 1
f(x) = \frac{1}{1 + x^2}
\triangle x = \frac{1-0}{n} = \frac{1}{n}
In your problem, the index of summation is called r instead of k so you see \frac{r}{n} in place of k \triangle x.

 

[andyrk]

So if I use the equation \alpha_n - \beta_n = \frac{1}{n}( f(1) - f(0) ) and not \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n) ), I get the answer of \frac{\alpha_n}{\alpha} - \frac{\beta_n}{\beta} as \frac{-2}{nπ} (i.e k, where k is some negative real number) and \alpha_n - \beta_n comes out to be \frac{-1}{2n}. Is that correct?

Also, the reason which I came up with for α = β = π/4. First of all
\alpha = \lim_{n\rightarrow \infty} \alpha_n => \alpha = \int_a^b f(x) dx
and
\beta = \lim_{n\rightarrow \infty} \beta_n => \beta = \int_a^b f(x) dx
When we evaluate the above integral, it comes out to be π/4. Now the question is, why is α = β? α and β are equal because their integrals come out to be the same. But why do the two integrals come out to be the same even though the limits are different? So question reduces down to that why is \lim_{n\rightarrow \infty}(\alpha_n) = \lim_{n\rightarrow \infty} (\beta_n)? The reason which I could think of as to why the limits are equal is because when n→∞, the difference in areas under the two curves for \alpha_n and \beta_n can be neglected and they can be said to be approximately equal. They are approximately equal because when we look closely, we realize that for \alpha_n, the area under the curve is measured from x = \frac{1}{n} to x = 1. But for \beta_n, the area under the curve is measured from x = \frac{0}{n} to x = \frac{n-1}{n}. This difference in area becomes negligible when n→∞ and so both the areas are measured from x = 0 to x = 1. This is because, for n→∞, \frac{1}{n} → 0 and \frac{n-1}{n} → 1.

As a result, α = β. Now, even if we use the equation \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n) ), we get the value of \alpha_n - \beta_n as \frac{-n}{1+n^2} and hence \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} = \frac{-n/(1+n^2)}{π/4} = \frac{-4n}{π(1+n^2)}( i.e k, where k is some negative real number).

if we use the equation \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n), just the value of \alpha_n - \beta_n changes from \frac{-1}{2n} to \frac{-n}{1+n^2}. The answer to the whole problem still remains the same, i.e, \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} still comes out to be a number k which is a negative real number, regardless of the equation used. So \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} = \frac{\alpha_n}{\alpha} - \frac{\beta_n}{\alpha} = \frac{\alpha_n - \beta_n}{\alpha} = \frac{-1/2n}{π/4} = \frac{-2}{nπ} = k, (where k is a negative real number).

So, all in all, if we use the equation \alpha_n - \beta_n = \frac{1}{n}( f(1) - f(0) ), then \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} = \frac{-2}{nπ} = k.

And if we use the equation \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n) ), then \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} = \frac{-4n}{π(1+n^2)} = k.

But in both the cases, k is a negative real number. So my point is that, even we use the wrong equations we get the right answer (pure co-incidence though). So the problem gets solved either way, but the right way should be known.

 

[Stephen Tashi]

and \alpha_n comes out to be π/4 - 1/2n = π/4 since n→∞. Is that correct?

Check you statement of the problem in post #10 and see if there is any justification for taking the limit of \alpha_n as n \rightarrow \infty

Does the problem ask the value of \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha}?
Or does it ask the value of \lim_{n \rightarrow \infty}(\frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} ) ?
Did you omit the limt?

 

[andyrk]

Check you statement of the problem in post #10 and see if there is any justification for taking the limit of \alpha_n as n \rightarrow \infty

Does the problem ask the value of \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha}?
Or does it ask the value of \lim_{n \rightarrow \infty}(\frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} ) ?
Did you omit the limt?

See my edited message. What you quoted, I changed it just a moment ago.
And the problem asks for the value of \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha}.

 

[Stephen Tashi]

I must go visit someone now. I'll get back to this in about 4 hours.

 

[andyrk]

I must go visit someone now. I'll get back to this in about 4 hours.

Sure. :)
Read #34 once you come back. It has been edited a lot.

The thing to understand from the question was that the summations were not equal. But when the same summations were converted to integrals by the means of a limit, then the integrals became equal. So the area when represented in the form of a summation is not equal. But when the same summations are converted into integrals then the area represented by the definite integral is equal. (For this question only, that is). Am I right?

 

[Stephen Tashi]

So if I use the equation \alpha_n - \beta_n = \frac{1}{n}( f(1) - f(0) ) and not \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n) ), I get the answer of \frac{\alpha_n}{\alpha} - \frac{\beta_n}{\beta} as \frac{-2}{nπ} (i.e k, where k is some negative real number) and \alpha_n - \beta_n comes out to be \frac{-1}{2n}. Is that correct?

Yes.

Also, the reason which I came up with for α = β = π/4. First of all
\alpha = \lim_{n\rightarrow \infty} \alpha_n => \alpha = \int_a^b f(x) dx
and
\beta = \lim_{n\rightarrow \infty} \beta_n => \beta = \int_a^b f(x) dx

You should write "=" instead of "=>".

When we evaluate the above integral, it comes out to be π/4. Now the question is, why is α = β? α and β are equal because their integrals come out to be the same

I'd prefer to say "because they result from taking limits of sums that approximate the same integral".

But why do the two integrals come out to be the same even though the limits are different? So question reduces down to that why is \lim_{n\rightarrow \infty}(\alpha_n) = \lim_{n\rightarrow \infty} (\beta_n)? The reason which I could think of as to why the limits are equal is because when n→∞, the difference in areas under the two curves for \alpha_n and \beta_n can be neglected and they can be said to be approximately equal.

Yes.

They are approximately equal because when we look closely, we realize that for \alpha_n, the area under the curve is measured from x = \frac{1}{n} to x = 1. But for \beta_n, the area under the curve is measured from x = \frac{0}{n} to x = \frac{n-1}{n}. This difference in area becomes negligible when n→∞ and so both the areas are measured from x = 0 to x = 1. This is because, for n→∞, \frac{1}{n} → 0 and \frac{n-1}{n} → 1.

That's a correct intuitive explanation. To prove formally that the areas are the same is harder, but that proof should haven been given in your text materials when you were introduced to definite integrals.

As a result, α = β. Now, even if we use the equation \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n) ), we get the value of \alpha_n - \beta_n as \frac{-n}{1+n^2} and hence \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} = \frac{-n/(1+n^2)}{π/4} = \frac{-4n}{π(1+n^2)}( i.e k, where k is some negative real number).

if we use the equation \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n), just the value of \alpha_n - \beta_n changes from \frac{-1}{2n} to \frac{-n}{1+n^2}. The answer to the whole problem still remains the same, i.e, \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} still comes out to be a number k which is a negative real number, regardless of the equation used. So \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} = \frac{\alpha_n}{\alpha} - \frac{\beta_n}{\alpha} = \frac{\alpha_n - \beta_n}{\alpha} = \frac{-1/2n}{π/4} = \frac{-2}{nπ} = k, (where k is a negative real number).

It will get confusing to discuss the results from two different problems. If you change the definitions of \alpha_n and \beta_n then you might have to change the limits of integration and the value of the integral would be different.

So, all in all, if we use the equation \alpha_n - \beta_n = \frac{1}{n}( f(1) - f(0) ), then \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} = \frac{-2}{nπ} = k.

Yes.

And if we use the equation \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n) )

Let's not worry about that!

 

[andyrk]

Thanks! Nice problem though.