Nice-Calculus-Find-the-value-of-n-1-1-4-n-cos-2-pi-2-n-2-

Question Number 144311 by mnjuly1970 last updated on 24/Jun/21

\dots \dots Nice \dots . Calculus \dots \dots

Find the value of ::

Θ := \underset{n = 1}{\sum^{\infty}} \frac{1}{4^{n} {c o s}^{2} (\frac{π}{2^{n + 2}})} = ?

\dots \dots \dots .

Commented by Kamel last updated on 24/Jun/21

S_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{4^{k} {c o s}^{2} (\frac{π}{2^{k + 2}})}, T_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{4^{k} {s i n}^{2} (\frac{π}{2^{k + 2}})}

S o : S_{n} + T_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{4^{k} {(s i n (\frac{π}{2^{k + 2}}) c o s (\frac{π}{2^{k + 2}}))}^{2}} = \underset{k = 1}{\sum^{n}} \frac{1}{4^{k - 1} {s i n}^{2} (\frac{π}{2^{k + 1}})}

\overset{p = k - 1}{=} \underset{p = 0}{\sum^{n - 1}} \frac{1}{4^{p} {s i n}^{2} (\frac{π}{2^{p + 2}})} = 2 + T_{n} - \frac{1}{4^{n} {s i n}^{2} (\frac{π}{2^{n + 2}})}

T h e n : S_{n} = 2 - \frac{1}{4^{n} {s i n}^{2} (\frac{π}{2^{n + 2}})}

∴ \underset{n = 1}{\sum^{+ \infty}} \frac{1}{4^{n} {c o s}^{2} (\frac{π}{2^{n + 2}})} = 2 - \underset{n \to + \infty}{l i m} {(\frac{\overset{}{1}}{2^{n} s i n (\frac{π}{2^{n + 2}})})}^{2} \overset{t = \frac{π}{2^{n + 2}}}{=} 2 - \underset{t \to 0^{+}}{l i m} {(\frac{4}{π} \frac{t}{s i n (t)})}^{2}

∴ \underset{n = 1}{\sum^{+ \infty}} \frac{1}{4^{n} {c o s}^{2} (\frac{π}{2^{n + 2}})} = 2 - \frac{16}{π^{2}}

Commented by mnjuly1970 last updated on 24/Jun/21

b r a v o m r k a m e l t a s h a k o r . .

Answered by ArielVyny last updated on 24/Jun/21

Θ = \underset{n = 1}{\sum^{\infty}} \frac{1}{4^{n}} [1 + {t g}^{2} (\frac{π}{2^{n + 2}})] = \underset{n = 1}{\sum^{\infty}} {(\frac{1}{4})}^{n} + \underset{n = 1}{\sum^{\infty}} \frac{{t g}^{2} (\frac{π}{2^{n + 2}})}{4^{n}}

Θ = \frac{4}{3} + \underset{n = 1}{\sum^{\infty}} {(\frac{1}{4})}^{n} {t g}^{2} (\frac{π}{4 \times 2^{n}})

p o s o n s {(\frac{1}{2})}^{n} = t

Θ = \frac{4}{3} + \underset{t = \frac{1}{2}}{\sum^{0}} t^{2} {t g}^{2} (\frac{π}{4} t) = \frac{4}{3} - \underset{t = 0}{\sum^{\frac{1}{2}}} t^{2} {t g}^{2} (\frac{π}{4} t)

Θ = \frac{4}{3} - \frac{1}{2} \underset{t = 0}{\sum^{1}} t^{2} {t g}^{2} (\frac{π}{4} t)

Θ = \frac{4}{3} - \frac{1}{2}

Θ = \frac{5}{6}

Commented by mathmax by abdo last updated on 25/Jun/21

not correct!

Answered by abdullahoudou last updated on 24/Jun/21

1 - {(2 / π)}^{2}

Commented by ArielVyny last updated on 24/Jun/21

y o u c a n s h o w t h a t p l e a s e ?