Integral on Sequence: Proving Summation Equation

In summary, the given equation reduces to a summation involving trigonometric functions and can be solved by splitting into partial fractions. By manipulating the terms and recognizing a pattern, it can be shown that the equation is equal to the summation of \frac{1}{n^2} from n=2 to infinity.
  • #1
burak100
33
0

Homework Statement


show that holds,

[itex] \sum\limits_{n=1}^{\infty}\int\limits_{0}^{\frac{\pi}{2}}\frac{(2n-1) sin(2n-1) x}{n^2(n+1)}dx = \sum\limits_{n=2}^{\infty}\frac{1}{n^2}[/itex]

Homework Equations





The Attempt at a Solution



Actually, I have no idea how should I start.
 
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  • #2
This reduces to

[tex]\sum_{n=1}^{+\infty}{\frac{2n-1}{n^2(n+1)}\int_0^{\pi/2}{\sin((2n-1)x)dx}}[/tex]

Now, calculate the integral.
 
  • #3
thank you very much for your help, I calculate the integral;

[itex]= \sum\limits_{n=1}^{\infty}\frac{(2n-1)}{n^2(n+1)}\Big( \frac{-1}{(2n-1)} \cos((2n-1)x) \vert_{0}^{\frac{\Pi}{2}} \Big)[/itex]

[itex]= \sum\limits_{n=1}^{\infty}\frac{-1}{n^2(n+1)} \Big( cos((2n-1)\frac{\Pi}{2}) -1 \Big)[/itex]

[itex]= \sum\limits_{n=1}^{\infty} \frac{1- cos((2n-1)\frac{\Pi}{2}) }{n^2(n+1)}[/itex]

for [itex]n=1[/itex], this summation becomes [itex]\frac{1}{2}[/itex]

so,

[itex]\frac{1}{2} +\sum\limits_{n=2}^{\infty} \frac{1- cos((2n-1)\frac{\Pi}{2}) }{n^2(n+1)}[/itex]

it becomes like that. but here how should I continue to show that this equation equal to

[itex] \sum\limits_{n=2}^{\infty} \frac{1}{n^2}[/itex]

??
 
  • #4
burak100 said:
thank you very much for your help, I calculate the integral;

[itex]= \sum\limits_{n=1}^{\infty}\frac{(2n-1)}{n^2(n+1)}\Big( \frac{-1}{(2n-1)} \cos((2n-1)x) \vert_{0}^{\frac{\Pi}{2}} \Big)[/itex]

[itex]= \sum\limits_{n=1}^{\infty}\frac{-1}{n^2(n+1)} \Big( cos((2n-1)\frac{\Pi}{2}) -1 \Big)[/itex]

[itex]= \sum\limits_{n=1}^{\infty} \frac{1- cos((2n-1)\frac{\Pi}{2}) }{n^2(n+1)}[/itex]

OK, but [itex]\cos((2n-1)\pi/2)[/itex] can be calculated, right??

for [itex]n=1[/itex], this summation becomes [itex]\frac{1}{2}[/itex]

so,

[itex]\frac{1}{2} + \frac{1- cos((2n-1)\frac{\Pi}{2}) }{n^2(n+1)}[/itex]

it becomes like that. but here how should I continue to show that this equation equal to

[itex] \sum\limits_{n=2}^{\infty} \frac{1}{n^2}[/itex]

??

I have no idea why you drop the summation here...
 
  • #5
Ok if I rewrite it, finally it becomes
[itex]\frac{}{}[/itex]
[itex]\[=\sum\limits_{n=1}^{\infty}\frac{1- \cos((2n-1)\frac{\Pi}{2})}{n^2(n+1)}\][/itex]
[itex]\[=\sum\limits_{n=1}^{\infty}\frac{1- \cos(n \Pi - \frac{\Pi}{2})}{n^2(n+1)}\][/itex]
[itex]\[=\sum\limits_{n=1}^{\infty}\frac{1- \Big(\cos(n\Pi)\cos(\frac{\Pi}{2}) + \sin(n\Pi)\sin(\frac{\Pi}{2})\Big)}{n^2(n+1)}\][/itex]
[itex]\[=\sum\limits_{n=1}^{\infty}\frac{1}{n^2(n+1)}\][/itex]

actually I can't go on ..

how it is equal to
[itex]\[=\sum\limits_{n=2}^{\infty}\frac{1}{n^2}\][/itex]
?
 
  • #6
OK, now split

[tex]\frac{1}{n^2(n+1)}[/tex]

into partial fractions...
 
  • #7
ok now I think we found, can you check,

[itex]=\sum\limits_{n=1}^{\infty}\frac{1}{n^2(n+1)}[/itex]
[itex]=\sum\limits_{n=1}^{\infty}\frac{1}{n^2} - \sum\limits_{n=1}^{\infty}\frac{1}{n(n+1)}[/itex]
[itex]=\sum\limits_{n=1}^{\infty}\frac{1}{n^2} - \Big( \sum\limits_{n=1}^{\infty}\frac{1}{n}- \sum\limits_{n=1}^{\infty}\frac{1}{(n+1)} \Big)[/itex]
[itex]=\sum\limits_{n=1}^{\infty}\frac{1}{n^2} - 1[/itex]
[itex]=\sum\limits_{n=2}^{\infty}\frac{1}{n^2}[/itex]

thank you very much...
 
  • #8
burak100 said:
[itex]=\sum\limits_{n=1}^{\infty}\frac{1}{n^2} - \Big( \sum\limits_{n=1}^{\infty}\frac{1}{n}- \sum\limits_{n=1}^{\infty}\frac{1}{(n+1)} \Big)[/itex]

Nonono, you can't write that. [itex]\sum_{n=1}^{+\infty}{\frac{1}{n}}[/itex] is infinity. So what's standing there is infinity - infinity. This doesn't make any sense.

You'll need another way to calculate

[tex]\sum_{n=1}^{+\infty}{\frac{1}{n}-\frac{1}{n+1}}[/tex]

Write out the first 10 terms of the sum and see if you notice anything...
 
  • #9
[itex]=\sum\limits_{n=1}^{\infty}\frac{1}{n(n+1)}[/itex]
[itex]=\sum\limits_{n=1}^{\infty}\Big( \frac{1}{n} - \frac{1}{n+1} \Big)[/itex]
[itex]= \Big( \frac{1}{1} - \frac{1}{2} \Big)[/itex]
[itex]~~+ \Big( \frac{1}{2} - \frac{1}{3} \Big)[/itex]
[itex]~~+ \Big( \frac{1}{3} - \frac{1}{4} \Big)[/itex]
[itex]~~+ \Big( \frac{1}{4} - \frac{1}{5} \Big)[/itex]
...
and so on,
then it is equal to [itex] 1 [/itex]
..
 
  • #10
Yes, that is the correct reasoning!
 
  • #11
Thank you very much..
 

1. What is an integral on sequence?

An integral on sequence is a mathematical concept that involves finding the area under a curve on a sequence of numbers. It is a way to express the sum of a sequence as a continuous function.

2. How do you prove the summation equation for an integral on sequence?

The summation equation for an integral on sequence can be proved using the fundamental theorem of calculus. This involves showing that the function representing the sum of the sequence is the derivative of the integral function.

3. What is the significance of the summation equation in integral on sequence?

The summation equation is significant because it allows us to calculate the sum of a sequence without having to manually add up each term. It also provides a way to find the sum of infinite sequences, which would be impossible to calculate otherwise.

4. Can the summation equation be used for any sequence?

Yes, the summation equation for an integral on sequence can be used for any sequence as long as it meets certain criteria. The sequence must be continuous and have a well-defined integral function.

5. How is the summation equation related to other mathematical concepts?

The summation equation is closely related to other mathematical concepts, such as the Riemann sum and the definite integral. It is also connected to the concept of convergence and divergence of infinite series.

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