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

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

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

diverges

Solution steps

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

Apply Series Comparison Test:

diverges

= diverges

n = 1 \sum \infty \frac{1}{( n + 1 ) n}

n = 1 \sum \infty sin (9)

n = 1 \sum \infty \frac{2}{7 ^{n}}

n = 0 \sum \infty (- 1 + \frac{4}{n})^{4}

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

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