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

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

n = 1 \sum \infty sin (2 n)

diverges

Solution steps

n = 1 \sum \infty sin (2 n)

Apply Series Divergence Test:

diverges

= diverges

\int \frac{1}{x ( x ^{2} - 25 ) ^{\frac{3}{2}}} d x

y = 2 x^{4} 25 - x^{2}

(ln (1 + x^{2}))^{^{'}}

\int (\frac{6}{x ^{3}}) d x

\frac{\partial}{\partial y} (6 x y^{2} - 6 x^{2} y + 5 x y)

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