n-1-1-2-n-tan-1-2-n-

Question Number 90637 by Tony Lin last updated on 25/Apr/20

\underset{n = 1}{\sum^{\infty}} \frac{1}{2^{n}} t a n (\frac{1}{2^{n}})

Answered by TANMAY PANACEA. last updated on 25/Apr/20

t a n θ - c o t θ

= \frac{{t a n}^{2} θ - 1}{t a n θ}

= - 2 (\frac{1 - {t a n}^{2} θ}{2 t a n θ}) = \frac{- 2}{t a n 2 θ}

t a n θ = c o t θ - 2 c o t 2 θ

S = \frac{1}{2} t a n (\frac{1}{2}) + \frac{1}{2^{2}} t a n (\frac{1}{2^{2}}) + \frac{1}{2^{3}} t a n (\frac{1}{2^{3}}) + \dots \infty

T_{1} = \frac{1}{2} t a n (\frac{1}{2}) = \frac{1}{2} c o t (\frac{1}{2}) - \frac{1}{2} \times 2 c o t (2 \times \frac{1}{2})

T_{2} = \frac{1}{2^{2}} t a n (\frac{1}{2^{2}}) = \frac{1}{2^{2}} c o t (\frac{1}{2^{2}}) - \frac{1}{2^{2}} \times 2 c o t (2 \times \frac{1}{2^{2}})

T_{3} = \frac{1}{2^{3}} t a n (\frac{1}{2^{3}}) = \frac{1}{2^{3}} c o t (\frac{1}{2^{3}}) - \frac{1}{2^{3}} \times 2 c o t (2 \times \frac{1}{2^{3}})

\dots .

\dots .

T_{n} = \frac{1}{2^{n}} t a n (\frac{1}{2^{n}}) = \frac{1}{2^{n}} c o t (\frac{1}{2^{n}}) - \frac{1}{2^{n}} \times 2 c o t (2 \times \frac{1}{2^{n}})

n o w a d d t h e m

S_{n} = \frac{1}{2^{n}} c o t (\frac{1}{2^{n}}) - \frac{1}{2} \times 2 c o t (2 \times \frac{1}{2})

Commented by Tony Lin last updated on 26/Apr/20

t h a n k s s i r

Commented by Tony Lin last updated on 26/Apr/20

t h e n

\lim_{n \to \infty} \frac{c o t (\frac{1}{2^{n}})}{2^{n}}

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

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

l e t \frac{1}{2^{n}} = t, n \to \infty, t \to 0

\Rightarrow \lim_{t \to 0} \frac{t}{s i n t}

= 1

t h e r e f o r e,

\underset{n = 1}{\sum^{\infty}} \frac{1}{2^{n}} t a n (\frac{1}{2^{n}})

= 1 - c o t 1

\approx 0.3579

Commented by TANMAY PANACEA. last updated on 26/Apr/20

t h a n k y o u s i r \dots w h e r e i s t o p p e d \dots y o u h e l p e d

t o r e a c h t h e g o a l \dots