calculus-2-evaluate-n-0-2-n-1-2-2-n-

Question Number 128489 by mnjuly1970 last updated on 07/Jan/21

\dots c a l c u l u s (2) \dots

e v a l u a t e ::

\underset{n = 0}{\sum^{\infty}} (\frac{2^{n}}{1 + 2^{2^{n}}}) = ?

Answered by mindispower last updated on 07/Jan/21

\frac{1}{1 - x^{2}} = \frac{1}{2} (\frac{1}{1 - x} + \frac{1}{1 + x})

\Rightarrow \frac{2}{1 - x^{2}} - \frac{1}{1 - x} = \frac{1}{1 + x}

x = 2^{2^{n}} \Rightarrow x^{2} = 2^{2^{n + 1}}

\Leftrightarrow \frac{2}{1 - 2^{2^{n + 1}}} - \frac{1}{1 - 2^{2^{n}}} = \frac{1}{1 + 2^{2^{n}}}

\Leftrightarrow \frac{2 . 2^{n}}{1 - 2^{2^{n + 1}}} - \frac{2^{n}}{1 - 2^{2^{n}}} = \frac{2^{n}}{1 + 2^{2^{n}}}

\Leftrightarrow \frac{2^{n + 1}}{1 - 2^{2^{n + 1}}} - \frac{2^{n}}{1 - 2^{2^{n}}} = \frac{2^{n}}{1 + 2^{2^{n}}}

l e t V_{n} = \frac{2^{n}}{1 - 2^{2^{n}}}

\frac{2^{n}}{1 + 2^{2^{n}}} = V_{n + 1} - V_{n}

\underset{n ⩾ 0}{\sum^{m}} \frac{2^{n}}{1 + 2^{2^{n}}} = \underset{n ⩾ 0}{\sum^{m}} V_{n + 1} - V_{n} = V_{m + 1} - V_{0} \dots E

\lim_{m \to \infty} \frac{2^{m}}{1 + 2^{2^{m}}} = \lim_{x \to \infty} \frac{x}{1 + 2^{x}} \to 0

\Rightarrow \lim_{m \to \infty} \underset{n ⩾ 0}{\sum^{m}} \frac{2^{n}}{1 + 2^{2^{n}}} = \sum_{n ⩾ 0} \frac{2^{n}}{1 + 2^{2^{n}}} = - V_{0} = - \frac{1}{1 - 2^{1}} = 1

S = 1

Commented by mnjuly1970 last updated on 08/Jan/21

n i c e v e r y n i c e . .