What is the sum from k=1 to infinity of (20)^k ?

The sum from k=1 to infinity of (20)^k is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from k=1 to infinity of (20)^k

k = 1 \sum \infty (20)^{k}

diverges

Solution steps

k = 1 \sum \infty 2 0^{k}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= k = 0 \sum \infty 2 0^{k} - 2 0^{0}

k = 0 \sum \infty 2 0^{k} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{- 4}^{4} ∣ x - 3 ∣ d x

f (x) = (\frac{x ^{3}}{x ^{8} + 5})^{2}

\frac{d y}{d x} = x^{\frac{1}{4}} + x^{- \frac{3}{5}}

\int_{\frac{1}{3}}^{5} 10 x ln (3 x) d x

\int_{0}^{2 π} x^{2} cos (n x) d x

What is the sum from k=1 to infinity of (20)^k ?
The sum from k=1 to infinity of (20)^k is diverges