What is the sum from n=1 to infinity of (3n)/(2n+1) ?

The sum from n=1 to infinity of (3n)/(2n+1) 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 (3n)/(2n+1)

n = 1 \sum \infty \frac{3 n}{2 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{3 n}{2 n + 1}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{n}{2 n + 1}

Apply Series Divergence Test:

diverges

= 3 diverges

= diverges

\frac{\partial}{\partial x} (\frac{y}{x ^{2} + y ^{2} + 1})

6 x y^{^{'}} = 2 x e^{x} - 6 y + 6 x^{2}

\frac{\partial}{\partial v} (u^{2})

\int \frac{1}{2 k ^{2}} \cdot e^{- \frac{x}{2 k ^{2}}} d x

\int \frac{1 - u}{1 + u ^{2}} d u

What is the sum from n=1 to infinity of (3n)/(2n+1) ?
The sum from n=1 to infinity of (3n)/(2n+1) is diverges