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

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

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

diverges

Solution steps

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

Apply Series Limit Comparison Test:

diverges

= diverges

\int_{- π}^{π} x^{2} cos (2 x) d x

\int sin^{2} (7 x) cos^{2} (7 x) d x

6 (5)^{2} - 2 (5) + 9

\frac{d}{d x} (e^{2 x} + x^{2} - 16)

\int_{0}^{2} (1 + 3 x^{2}) d x

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