What is the sum from n=1 to infinity of (n^2)/n ?

The sum from n=1 to infinity of (n^2)/n 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^2)/n

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

x = 2 \sum \infty \frac{1}{x ln ( x )}

n = 1 \sum \infty cos (n \cdot π)

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

n = 1 \sum \infty cos (n) π

n = 1 \sum \infty \frac{2 n}{9 n + 1}

What is the sum from n=1 to infinity of (n^2)/n ?
The sum from n=1 to infinity of (n^2)/n is diverges