What is the sum from n=1 to infinity of 1/(n^2+300) ?

The sum from n=1 to infinity of 1/(n^2+300) is converges

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 1/(n^2+300)

n = 1 \sum \infty \frac{1}{n ^{2} + 300}

converges

Solution steps

n = 1 \sum \infty \frac{1}{n ^{2} + 300}

Apply Series Comparison Test:

converges

= converges

y^{^{'}} = 2 y (5 - y)

f (x) = (x^{2} - 2) (x^{3} - 3 x), at x = - 2

x \to 0 lim (\frac{x ^{2} - 5}{x - 2})

\frac{d ^{2}}{d x ^{2}} x y + 2 x^{2} = 5

\int sin^{3} (x) \cdot cos (x) d x

What is the sum from n=1 to infinity of 1/(n^2+300) ?
The sum from n=1 to infinity of 1/(n^2+300) is converges