What is the sum from n=1 to infinity of 1/(4n+2) ?

The sum from n=1 to infinity of 1/(4n+2) 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 1/(4n+2)

n = 1 \sum \infty \frac{1}{4 n + 2}

diverges

Solution steps

n = 1 \sum \infty \frac{1}{4 n + 2}

Apply Series Integral Test:

diverges

= diverges

7 y ln (x) - x y^{^{'}} = 0

\frac{d}{d x} (\frac{x}{4 - x ^{2}})

g (u) = 3 u + 5 u

f (x) = 15 e^{x} + 11 x, at x = 0

\int cos^{2} (4 x) sin (4 x) d x

What is the sum from n=1 to infinity of 1/(4n+2) ?
The sum from n=1 to infinity of 1/(4n+2) is diverges