What is the sum from n=1 to infinity of (3-2n)/(4n) ?

The sum from n=1 to infinity of (3-2n)/(4n) 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 (3-2n)/(4n)

n = 1 \sum \infty \frac{3 - 2 n}{4 n}

diverges

Solution steps

n = 1 \sum \infty \frac{3 - 2 n}{4 n}

Apply the constant multiplication rule:

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

= \frac{1}{4} \cdot n = 1 \sum \infty \frac{3 - 2 n}{n}

Apply Series Divergence Test:

diverges

= \frac{1}{4} diverges

= diverges

(- 2, 7), (2, 3)

\int sin^{4} (a x) d x

\frac{d}{d x} (10 x e^{- 2 x})

y = x^{2} - 4, y = 3 x

\frac{d}{d x} (\frac{4 x ^{2} - \frac{1}{2} x}{9 x + 8})

What is the sum from n=1 to infinity of (3-2n)/(4n) ?
The sum from n=1 to infinity of (3-2n)/(4n) is diverges