What is the sum from n=1 to infinity of 1/(n^{0.8)} ?

The sum from n=1 to infinity of 1/(n^{0.8)} 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 1/(n^{0.8)}

n = 1 \sum \infty \frac{1}{n ^{0.8}}

diverges

Solution steps

n = 1 \sum \infty \frac{1}{n ^{0.8}}

Apply p-Series Test:

diverges

= diverges

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

n = 1 \sum \infty (\frac{6 π}{7})^{n}

n = 4 \sum \infty \frac{1}{n ^{2} - n}

n = 0 \sum \infty (- 9)^{n}

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

What is the sum from n=1 to infinity of 1/(n^{0.8)} ?
The sum from n=1 to infinity of 1/(n^{0.8)} is diverges