What is the sum from n=1 to infinity of n^{-0.6} ?

The sum from n=1 to infinity of n^{-0.6} is diverges

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 n^{-0.6}

n = 1 \sum \infty n^{- 0.6}

diverges

Solution steps

n = 1 \sum \infty n^{- 0.6}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= n = 1 \sum \infty \frac{1}{n ^{0.6}}

Apply p-Series Test:

diverges

= diverges

n = 0 \sum \infty \frac{4}{9 - n ^{2}}

k = 1 \sum \infty \frac{3}{k} - \frac{3}{k + 2}

n = 1 \sum \infty \frac{6 n}{6 n + 1}

n = 1 \sum \infty \frac{1}{n} - \frac{2}{3 n - 1}

k = 2 \sum \infty (- 0.95)^{k}

What is the sum from n=1 to infinity of n^{-0.6} ?
The sum from n=1 to infinity of n^{-0.6} is diverges