What is the sum from n=1 to infinity of-1/(2^n) ?

The sum from n=1 to infinity of-1/(2^n) is -1

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/(2^n)

n = 1 \sum \infty - \frac{1}{2 ^{n}}

- 1

Solution steps

n = 1 \sum \infty - \frac{1}{2 ^{n}}

Apply the constant multiplication rule:

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

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

n = 0 \sum \infty \frac{1}{2 ^{n}} = 2

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

Simplify

= - 1

n = 0 \sum \infty \frac{6}{n}

i = 1 \sum \infty e^{- i}

n = 1 \sum \infty \frac{( - 3 + 2 ) ^{n}}{n}

n = 0 \sum \infty \frac{2}{1 + 3 n}

n = 2 \sum \infty \frac{1}{n + 2}

What is the sum from n=1 to infinity of-1/(2^n) ?
The sum from n=1 to infinity of-1/(2^n) is -1