What is the sum from n=0 to infinity of (3n)/(4n-1) ?

The sum from n=0 to infinity of (3n)/(4n-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=0 to infinity of (3n)/(4n-1)

n = 0 \sum \infty \frac{3 n}{4 n - 1}

diverges

Solution steps

n = 0 \sum \infty \frac{3 n}{4 n - 1}

Apply the constant multiplication rule:

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

= 3 \cdot n = 0 \sum \infty \frac{n}{4 n - 1}

Apply Series Divergence Test:

diverges

= 3 diverges

= diverges

y = arctan (ln (2) x^{5})

f (x) = 5 x^{5} + \frac{2}{x}

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

y = x^{2} - x + 9

\frac{d y}{( 5 - 3 y ) ^{2}} = \frac{d x}{2 ( 4 - x )}

What is the sum from n=0 to infinity of (3n)/(4n-1) ?
The sum from n=0 to infinity of (3n)/(4n-1) is diverges