What is the sum from n=1 to infinity of sin(1/(2^n)) ?

The sum from n=1 to infinity of sin(1/(2^n)) is converges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=1 to infinity of sin(1/(2^n))

n = 1 \sum \infty sin (\frac{1}{2 ^{n}})

converges

Solution steps

n = 1 \sum \infty sin (\frac{1}{2 ^{n}})

Apply Series Integral Test:

converges

= converges

2 \cdot cos (5 t)

\frac{\partial}{\partial y} (e^{5 x} sin (5 y))

\int - 4 tan (x) d x

\int 35 sec^{- 2} (x t a) n^{3} x d x

x \to 1 lim (\frac{1 - x}{x ^{2} - 1})

What is the sum from n=1 to infinity of sin(1/(2^n)) ?
The sum from n=1 to infinity of sin(1/(2^n)) is converges