# Integral on sequence

1. Sep 10, 2011

### burak100

1. The problem statement, all variables and given/known data
show that holds,

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

2. Relevant equations

3. The attempt at a solution

Actually, I have no idea how should I start.

2. Sep 10, 2011

### micromass

Staff Emeritus
This reduces to

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

Now, calculate the integral.

3. Sep 10, 2011

### burak100

thank you very much for your help, I calculate the integral;

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

$= \sum\limits_{n=1}^{\infty}\frac{-1}{n^2(n+1)} \Big( cos((2n-1)\frac{\Pi}{2}) -1 \Big)$

$= \sum\limits_{n=1}^{\infty} \frac{1- cos((2n-1)\frac{\Pi}{2}) }{n^2(n+1)}$

for $n=1$, this summation becomes $\frac{1}{2}$

so,

$\frac{1}{2} +\sum\limits_{n=2}^{\infty} \frac{1- cos((2n-1)\frac{\Pi}{2}) }{n^2(n+1)}$

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

$\sum\limits_{n=2}^{\infty} \frac{1}{n^2}$

???????

4. Sep 10, 2011

### micromass

Staff Emeritus
OK, but $\cos((2n-1)\pi/2)$ can be calculated, right??

I have no idea why you drop the summation here...

5. Sep 10, 2011

### burak100

Ok if I rewrite it, finally it becomes
$\frac{}{}$
$$=\sum\limits_{n=1}^{\infty}\frac{1- \cos((2n-1)\frac{\Pi}{2})}{n^2(n+1)}$$
$$=\sum\limits_{n=1}^{\infty}\frac{1- \cos(n \Pi - \frac{\Pi}{2})}{n^2(n+1)}$$
$$=\sum\limits_{n=1}^{\infty}\frac{1- \Big(\cos(n\Pi)\cos(\frac{\Pi}{2}) + \sin(n\Pi)\sin(\frac{\Pi}{2})\Big)}{n^2(n+1)}$$
$$=\sum\limits_{n=1}^{\infty}\frac{1}{n^2(n+1)}$$

actually I can't go on ..

how it is equal to
$$=\sum\limits_{n=2}^{\infty}\frac{1}{n^2}$$
????

6. Sep 10, 2011

### micromass

Staff Emeritus
OK, now split

$$\frac{1}{n^2(n+1)}$$

into partial fractions...

7. Sep 10, 2011

### burak100

ok now I think we found, can you check,

$=\sum\limits_{n=1}^{\infty}\frac{1}{n^2(n+1)}$
$=\sum\limits_{n=1}^{\infty}\frac{1}{n^2} - \sum\limits_{n=1}^{\infty}\frac{1}{n(n+1)}$
$=\sum\limits_{n=1}^{\infty}\frac{1}{n^2} - \Big( \sum\limits_{n=1}^{\infty}\frac{1}{n}- \sum\limits_{n=1}^{\infty}\frac{1}{(n+1)} \Big)$
$=\sum\limits_{n=1}^{\infty}\frac{1}{n^2} - 1$
$=\sum\limits_{n=2}^{\infty}\frac{1}{n^2}$

thank you very much....

8. Sep 10, 2011

### micromass

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

You'll need another way to calculate

$$\sum_{n=1}^{+\infty}{\frac{1}{n}-\frac{1}{n+1}}$$

Write out the first 10 terms of the sum and see if you notice anything...

9. Sep 10, 2011

### burak100

$=\sum\limits_{n=1}^{\infty}\frac{1}{n(n+1)}$
$=\sum\limits_{n=1}^{\infty}\Big( \frac{1}{n} - \frac{1}{n+1} \Big)$
$= \Big( \frac{1}{1} - \frac{1}{2} \Big)$
$~~+ \Big( \frac{1}{2} - \frac{1}{3} \Big)$
$~~+ \Big( \frac{1}{3} - \frac{1}{4} \Big)$
$~~+ \Big( \frac{1}{4} - \frac{1}{5} \Big)$
........
and so on,
then it is equal to $1$
..

10. Sep 10, 2011

### micromass

Staff Emeritus
Yes, that is the correct reasoning!!

11. Sep 10, 2011

### burak100

Thank you very much..