What is the sum from n=2 to infinity of (2/5)^n ?

The sum from n=2 to infinity of (2/5)^n is 4/15

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

sum from n=2 to infinity of (2/5)^n

n = 2 \sum \infty (\frac{2}{5})^{n}

\frac{4}{15}

Decimal

0.26666 \dots

Solution steps

n = 2 \sum \infty (\frac{2}{5})^{n}

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

= n = 0 \sum \infty (\frac{2}{5})^{n} - (\frac{2}{5})^{0} - (\frac{2}{5})^{1}

n = 0 \sum \infty (\frac{2}{5})^{n} = \frac{5}{3}

= \frac{5}{3} - (\frac{2}{5})^{0} - (\frac{2}{5})^{1}

Simplify

\frac{5}{3} - (\frac{2}{5})^{0} - (\frac{2}{5})^{1} : \frac{4}{15}

= \frac{4}{15}

What is the sum from n=2 to infinity of (2/5)^n ?
The sum from n=2 to infinity of (2/5)^n is 4/15