What is the sum from n=1 to infinity of (cos(14))^n ?

The sum from n=1 to infinity of (cos(14))^n is converges

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 (cos(14))^n

n = 1 \sum \infty (cos (14))^{n}

converges

Solution steps

n = 1 \sum \infty (cos (14))^{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 (cos (14))^{n} - (cos (14))^{0}

n = 0 \sum \infty (cos (14))^{n} =

converges

= converges - (cos (14))^{0}

= converges

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

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

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

k = 0 \sum \infty (0.4)^{k}

n = 1 \sum \infty n^{3} \frac{1}{n !}

What is the sum from n=1 to infinity of (cos(14))^n ?
The sum from n=1 to infinity of (cos(14))^n is converges