What is the sum from n=1 to infinity of (3/4)^{-n} ?

The sum from n=1 to infinity of (3/4)^{-n} is diverges

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 (3/4)^{-n}

n = 1 \sum \infty (\frac{3}{4})^{- n}

diverges

Solution steps

n = 1 \sum \infty (\frac{3}{4})^{- n}

Simplify

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

= n = 1 \sum \infty \frac{1}{( \frac{3}{4} ) ^{n}}

Simplify

\frac{1}{( \frac{3}{4} ) ^{n}} : (\frac{4}{3})^{n}

= n = 1 \sum \infty (\frac{4}{3})^{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{4}{3})^{n} - (\frac{4}{3})^{0}

n = 0 \sum \infty (\frac{4}{3})^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

What is the sum from n=1 to infinity of (3/4)^{-n} ?
The sum from n=1 to infinity of (3/4)^{-n} is diverges