What is the sum from n=1 to infinity of 3*(1/5)^n ?

The sum from n=1 to infinity of 3*(1/5)^n is 3/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

sum from n=1 to infinity of 3*(1/5)^n

n = 1 \sum \infty 3 \cdot (\frac{1}{5})^{n}

\frac{3}{4}

Decimal

0.75

Solution steps

n = 1 \sum \infty 3 (\frac{1}{5})^{n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty (\frac{1}{5})^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= 3 (n = 0 \sum \infty (\frac{1}{5})^{n} - (\frac{1}{5})^{0})

n = 0 \sum \infty (\frac{1}{5})^{n} = \frac{5}{4}

= 3 (\frac{5}{4} - (\frac{1}{5})^{0})

Simplify

3 (\frac{5}{4} - (\frac{1}{5})^{0}) : \frac{3}{4}

= \frac{3}{4}

What is the sum from n=1 to infinity of 3*(1/5)^n ?
The sum from n=1 to infinity of 3*(1/5)^n is 3/4