What is the sum from n=0 to infinity of n/(n^2+9) ?

The sum from n=0 to infinity of n/(n^2+9) 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=0 to infinity of n/(n^2+9)

n = 0 \sum \infty \frac{n}{n ^{2} + 9}

diverges

Solution steps

n = 0 \sum \infty \frac{n}{n ^{2} + 9}

Apply Series Integral Test:

diverges

= diverges

f (x) = \frac{x ^{2} + 4}{x}

\frac{d}{d x} (\frac{x ^{3} - x ^{2} + 1}{x - 2})

a x^{2}

\int_{- \infty}^{4} r e^{\frac{r}{2}} d r

\frac{d}{d x} (x^{2} + 5 x - 2)

What is the sum from n=0 to infinity of n/(n^2+9) ?
The sum from n=0 to infinity of n/(n^2+9) is diverges