What is the sum from n=0 to infinity of sin(1/n) ?

The sum from n=0 to infinity of sin(1/n) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=0 to infinity of sin(1/n)

n = 0 \sum \infty sin (\frac{1}{n})

diverges

Solution steps

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

Apply Series Limit Comparison Test:

diverges

= diverges

\int x^{2} 5 x + 8 d x

6 (x - 3)^{\frac{2}{3}}

\int_{1}^{9} \frac{1}{x - 5} d x

x \to 0 lim (x cos (\frac{1}{x}))

\int x y e^{x + y} d x

What is the sum from n=0 to infinity of sin(1/n) ?
The sum from n=0 to infinity of sin(1/n) is diverges