let-u-n-ln-cos-2-n-calculate-n-0-u-n-

Question Number 52678 by maxmathsup by imad last updated on 11/Jan/19

l e t u_{n} = l n {c o s (2^{- n})} c a l c u l a t e \sum_{n = 0}^{\infty} u_{n}

Commented by Abdo msup. last updated on 12/Jan/19

l e t S_{n} = \sum_{k = 0}^{n} u_{k} \Rightarrow S_{n} = \sum_{k = 0}^{n} l n (c o s (\frac{1}{2^{k}}))

= l n (\prod_{k = 0}^{n} c o s (\frac{1}{2^{k}})) l e t s i m p l i f y

W_{n} = \prod_{k = 0}^{n} c o s (\frac{1}{2^{k}}) l e t K_{n} = \prod_{k = 0}^{n} s i n (\frac{1}{2^{k}}) \Rightarrow

W_{n} . K_{n} = \prod_{k = 0}^{n} {c o s (\frac{1}{2^{k}}) s i n (\frac{1}{2^{k}})}

= c o s (1) s i n (1) \frac{1}{2^{n}} \prod_{k = 1}^{n} s i n (\frac{2}{2^{k}})

= \frac{1}{2} s i n (2) \prod_{k = 1}^{n} s i n (\frac{1}{2^{k - 1}})

= \frac{s i n (2)}{2} \prod_{k = 0}^{n - 1} s i n (\frac{1}{2^{k}}) = \frac{s i n (2)}{2} \frac{K_{n}}{s i n (\frac{1}{2^{n}})} \Rightarrow

W_{n} = \frac{s i n (2)}{2} \frac{1}{2^{n} s i n (\frac{1}{2^{n}})} b u t {l i m}_{n \to + \infty} \frac{1}{2^{n} s i n (\frac{1}{2^{n}})}

= {l i m}_{n \to + \infty} \frac{\frac{1}{2^{n}}}{s i n (\frac{1}{2^{n}})} = 1 ({l i m}_{x \to 0} \frac{x}{s i n x} = 1) \Rightarrow

{l i m}_{n \to + \infty} W_{n} = \frac{s i n 2}{2} \Rightarrow {l i m}_{n \to + \infty} S_{n} = l n (\frac{s i n 2}{2}) .

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jan/19

u_{0} + u_{1} + u_{2} + \dots + u_{n} + \dots \infty

= l n {c o s (\frac{1}{2^{0}})} + l n {c o s (\frac{1}{2^{1}})} + l n {c o s (\frac{1}{2^{2}})} + . . + l n {c o s (\frac{1}{2^{n - 1}})} + . . \infty

= l n {c o s (\frac{1}{2^{0}}) c o s (\frac{1}{2^{1}}) c o s (\frac{1}{2^{2}}) . . c o s (\frac{1}{2^{n - 1}}) \dots \infty} t

n o w

s i n (2) = 2 s i n 1 c o s 1

= 2^{2} s i n (\frac{1}{2}) (c o s (\frac{1}{2}) c o s 1

= 2^{3} s i n (\frac{1}{2^{2}}) c o s (\frac{1}{2^{2}}) c o s (\frac{1}{2}) c o s (\frac{1}{2^{0}})

t h u s r e q u i r e d a n s i s

l n {\frac{s i n 2}{2^{n} s i n (\frac{1}{2^{n - 1}}}} w h e n n \to \infty

\lim_{n \to 0 \infty} \frac{s i n (\frac{1}{2^{n - 1}})}{2^{n - 1}} \to 1

s o a n s w e r i s l n (\frac{s i n 2}{2})

a n o t h e r a p p r o a c h \dots

f o r m u l a

s i n θ = θ c o s (\frac{θ}{2}) c o s (\frac{θ}{2^{2}}) c o s (\frac{θ}{2^{3}}) \dots \infty

h e r e u_{n} = l n {c o s (\frac{1}{2^{n}})}

u_{0} + u_{1} + u_{2} + \dots \infty

l n {c o s (\frac{1}{2^{0}})} + l n {c o s (\frac{1}{2^{1}})} + l n {c o s (\frac{1}{2^{2}})} + \dots \infty

= l n {c o s (\frac{1}{2^{0}}) c o s (\frac{1}{2^{1}}) c o s (\frac{1}{2^{2}}) \dots \infty}

= l n {c o s (\frac{1}{2^{0}}) \times \frac{s i n 1}{1}} [p u t θ = 1]

= l n {\frac{2 s i n 1 c o s 1}{2}}

= l n (\frac{s i n 2}{2})

p l s c h e c k \dots .

Commented by Abdo msup. last updated on 12/Jan/19

t h a n k s s i r T a n m a y y o u r a n s w e r i s c o r r e c t .