What is the sum from n=1 to infinity of (n^4)/(n^5) ?

The sum from n=1 to infinity of (n^4)/(n^5) 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^4)/(n^5)

n = 1 \sum \infty \frac{n ^{4}}{n ^{5}}

diverges

Solution steps

n = 1 \sum \infty \frac{n ^{4}}{n ^{5}}

Simplify

\frac{n ^{4}}{n ^{5}} : \frac{1}{n}

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

Apply Cauchy's Convergence Condition:

diverges

= diverges

\frac{\partial}{\partial z} (2 x - 3 yz)

\int \frac{5 x - 4}{x ^{2} + 6 x + 13} d x

\int 4 x^{3} + 3 x^{2} + 2 x + 1 + \frac{1}{x} + \frac{1}{x ^{3}} d x

x^{2} y^{^{''}} - 2 x y^{^{'}} - 4 y = 0

What is the sum from n=1 to infinity of (n^4)/(n^5) ?
The sum from n=1 to infinity of (n^4)/(n^5) is diverges