What is the sum from n=1 to infinity of (1/11)^n ?

The sum from n=1 to infinity of (1/11)^n is 1/10

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 (1/11)^n

n = 1 \sum \infty (\frac{1}{11})^{n}

\frac{1}{10}

Solution steps

n = 1 \sum \infty (\frac{1}{11})^{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}{11})^{n} - (\frac{1}{11})^{0}

n = 0 \sum \infty (\frac{1}{11})^{n} = \frac{11}{10}

= \frac{11}{10} - (\frac{1}{11})^{0}

Simplify

\frac{11}{10} - (\frac{1}{11})^{0} : \frac{1}{10}

= \frac{1}{10}

\int x \cdot sin (a x) d x

n = 0 \sum \infty \frac{2 ^{n}}{100}

x^{2} y^{^{''}} + x y^{^{'}} - y = \frac{1}{x + 1}

x \to 1 - lim (\frac{x}{ln ( x )})

\int sin (x^{3} + 1) d x

What is the sum from n=1 to infinity of (1/11)^n ?
The sum from n=1 to infinity of (1/11)^n is 1/10