What is the sum from n=1 to infinity of 5(4)^{n-1} ?

The sum from n=1 to infinity of 5(4)^{n-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 5(4)^{n-1}

n = 1 \sum \infty 5 (4)^{n - 1}

diverges

Solution steps

n = 1 \sum \infty 5 \cdot 4^{n - 1}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty 4^{n - 1}

Apply Series Ratio Test:

diverges

= 5 diverges

= diverges

cos (3 x)

\frac{d}{d x} (\frac{1.05}{1 + 100 e ^{- 0.285 x}})

\int_{0}^{1.2} 1 - \frac{y}{2} d y

\frac{7 s}{s ^{2} - 2 s + 10}

\int \frac{t}{t + 1} d t

What is the sum from n=1 to infinity of 5(4)^{n-1} ?
The sum from n=1 to infinity of 5(4)^{n-1} is diverges