What is the sum from n=2 to infinity of 7/(4n+2) ?

The sum from n=2 to infinity of 7/(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=2 to infinity of 7/(4n+2)

n = 2 \sum \infty \frac{7}{4 n + 2}

diverges

Solution steps

n = 2 \sum \infty \frac{7}{4 n + 2}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 7 \cdot n = 2 \sum \infty \frac{1}{4 n + 2}

Apply Series Integral Test:

diverges

= 7 diverges

= diverges

y = x^{3} - 4 x + 4

\int \frac{y}{y ^{2}} d y

\frac{d}{d x} (x + \frac{400}{x})

\frac{5 s + 5.1}{s ^{2} + s + 20}

\frac{\partial}{\partial x} (x^{2} sin (10 t))

What is the sum from n=2 to infinity of 7/(4n+2) ?
The sum from n=2 to infinity of 7/(4n+2) is diverges