Integrated Summation

The summation clearly hints at the existence of a pattern among the sequence of integrals. Let us figure that out. Consider the following sum,

$\displaystyle a_n + a_{n+2} = \int\limits_0^\frac{\pi}{2} \frac{1- \cos 2nx }{1- \cos 2x } \text{d}x\ + \int\limits_0^\frac{\pi}{2} \frac{1- \cos 2(n+2)nx }{1- \cos 2x } \text{d}x$

$\displaystyle a_n + a_{n+2} = \int\limits_0^\frac{\pi}{2} \frac{2- (\cos 2nx + \cos 2(n+2)x)}{1- \cos 2x } \text{d}x$

$\displaystyle \Rightarrow a_n + a_{n+2} = \int\limits_0^\frac{\pi}{2} \frac{2- 2\cos 2(n+1)x \cos 2x}{1- \cos 2x } \text{d}x$

On simple rearrangement, we obtain,
$\displaystyle \Rightarrow a_n + a_{n+2} = \int\limits_0^\frac{\pi}{2} 2\cos 2(n+1)x \text{d}x + 2a_{n+1}$

$\displaystyle \Rightarrow a_n + a_{n+2} = 2\left.\frac{\sin 2(n+1)x}{2(n+1)}\right|_0^\frac{\pi}{2} + 2a_{n+1}$

$\displaystyle \Rightarrow a_n + a_{n+2} = 2a_{n+1}$ .

Hence it is an Arithmetic Progression with its initial term as $a_1$ and common difference as $a_2 - a_1$ . For this we need to evaluate $a_1$ and $a_2$ , which are very easy to evaluate. We get,

$\displaystyle a_1 = \frac{\pi}{2} , a_2 = \pi$ . Hence, common difference $= a_2 - a_1 = \frac{\pi}{2} = a_1$ .

Everything has now become very easy. $\displaystyle a_r = r a_1 \Rightarrow a_r a_{r+1} a_{r+2} a_{r+3} = r(r+1)(r+2)(r+3) \times a_1^4$

Therefore,
$\displaystyle \sum_{r=1}^n a_r a_{r+1} a_{r+2} a_{r+3} = a_1^4 \sum_{r=1}^n r(r+1)(r+2)(r+3) = \frac{ a_1^4}{5 } \times n(n+1)(n+2)(n+3)(n+4)$

Thus,
$\displaystyle S = \frac{5}{ a_1^4} \sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)(n+3)(n+4)}$

Now, the required sum is a standard telescopic series and the value of summation evaluates to

$\displaystyle S = \frac{5}{4\times 4!} \big( \frac{2}{\pi} \big)^4$

$\displaystyle \Rightarrow \boxed{24 \times \pi ^4 S = 20}$

Clearly, $a_{n} = \displaystyle \int_{0}^{\pi/2} \dfrac{\sin^2 nx}{\sin^2 x} dx = \displaystyle \int_{0}^{\pi/2} \sin^2 nx \csc^2 x dx$

Use IBP, using $\sin^2 nx$ as first part $\csc^2 x$ as second part to get :

$a_{n} = \displaystyle n \int_{0}^{\pi/2} \sin(2n x) \cot x dx$

= $n I(n)$ , where $I(n) = \displaystyle \int_{0}^{\pi/2} \sin(2n x) \cot x dx$

= $\displaystyle \int_{0}^{\pi/2} \dfrac{\sin(2n \frac{2x}{2})}{\sin(2x/2)} \cos(2(1-n)x + (2n-1) 2x/2) dx$

= $\displaystyle \int_{0}^{\pi/2} (\cos(2(1-n)x) + cos(2(2-n)x) + \dots + \cos(0 x) + \dots \cos 2n x) dx$

= $\dfrac{\pi}{2}$

Hence, $a_{n} = n I(n) = n \dfrac{\pi}{2}$

Rest is simple summing using telescoping series.

instead of thinking about generalisation $$ just integrate upto 3 times $$ taking n as 0, 1, 2 , you can easily see the recurrence is an A.P. of common difference $\pi$ /2 .... $$ and then again telescoping would take time just sum it upto 2 or 3 terms and $$ multiply as the change in sum would get enough small and then take it's ceiling function ! $$ you get 20 which is the answer , easy enough ! :)