What is the sum from n=1 to infinity of 1/(7^n) ?

The sum from n=1 to infinity of 1/(7^n) is 1/6

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 1/(7^n)

n = 1 \sum \infty \frac{1}{7 ^{n}}

\frac{1}{6}

Solution steps

n = 1 \sum \infty \frac{1}{7 ^{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{1}{7 ^{n}} - \frac{1}{7 ^{0}}

n = 0 \sum \infty \frac{1}{7 ^{n}} = \frac{7}{6}

= \frac{7}{6} - \frac{1}{7 ^{0}}

Simplify

\frac{7}{6} - \frac{1}{7 ^{0}} : \frac{1}{6}

= \frac{1}{6}

\int (x - 2)^{2} sin (x) d x

\int \frac{1}{x ^{2} + 3 x - 4} d x

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = - 7

\int sin (x) + sec^{2} (x) + sec (x) tan (x) d x

\int \frac{3}{x} + x^{3} d x

What is the sum from n=1 to infinity of 1/(7^n) ?
The sum from n=1 to infinity of 1/(7^n) is 1/6