What is the sum from n=1 to infinity of (n^2)/(5+2n) ?

The sum from n=1 to infinity of (n^2)/(5+2n) 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)/(5+2n)

n = 1 \sum \infty \frac{n ^{2}}{5 + 2 n}

diverges

Solution steps

n = 1 \sum \infty \frac{n ^{2}}{5 + 2 n}

Apply Series Divergence Test:

diverges

= diverges

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

k = 4 \sum \infty \frac{1}{k ^{2} - 5 k + 6}

n = 0 \sum \infty (- 1)^{n} (1)^{n}

n = 1 \sum \infty \frac{e ^{4 n}}{n !}

x = 1 \sum \infty x

What is the sum from n=1 to infinity of (n^2)/(5+2n) ?
The sum from n=1 to infinity of (n^2)/(5+2n) is diverges