# Sum of Series

1. Dec 7, 2014

### 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)

2. Dec 7, 2014

### andyrk

Is anyone there?

3. Dec 7, 2014

### 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.

4. Dec 7, 2014

### 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?

5. Dec 7, 2014

### 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.

6. Dec 7, 2014

### 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.

7. Dec 7, 2014

### Stephen Tashi

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

8. Dec 7, 2014

### andyrk

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.

9. Dec 7, 2014

### Stephen Tashi

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

10. Dec 8, 2014

### andyrk

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: Dec 8, 2014
11. Dec 8, 2014

### Stephen Tashi

Is the sum supposed to be $\sum_{r=0}^{n-1} f(\frac{r}{n})$ ? It begins at $r = 0$ ?

12. Dec 8, 2014

### andyrk

Yes. I didn't know how to write it up. Anyways, I have edited it now. :)

13. Dec 8, 2014

### 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 $\beta_n$

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

14. Dec 8, 2014

### andyrk

15. Dec 8, 2014

### 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.

16. Dec 8, 2014

### 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.

#### Attached Files:

• ###### Solution.jpg
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Last edited: Dec 8, 2014
17. Dec 8, 2014

### zoki85

No:D

18. Dec 8, 2014

### Stephen Tashi

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. Dec 8, 2014

### 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?

20. Dec 8, 2014

### Stephen Tashi

Have you studied definite integrals? Riemann sums?

21. Dec 8, 2014

### andyrk

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

22. Dec 8, 2014

### 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.

23. Dec 8, 2014

### 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)?

24. Dec 8, 2014

### 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 $k = n$.

25. Dec 8, 2014

### Stephen Tashi

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.