What is the sum from n=0 to infinity of 1/(9-n^2) ?

The sum from n=0 to infinity of 1/(9-n^2) is converges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

sum from n=0 to infinity of 1/(9-n^2)

n = 0 \sum \infty \frac{1}{9 - n ^{2}}

converges

Solution steps

n = 4 \sum \infty \frac{1}{9 - n ^{2}}

Factor

9 - n^{2} : - (- 9 + n^{2})

= n = 4 \sum \infty \frac{1}{- ( - 9 + n ^{2} )}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= - n = 4 \sum \infty \frac{1}{- 9 + n ^{2}}

Apply Series Integral Test:

converges

= - converges

= converges

What is the sum from n=0 to infinity of 1/(9-n^2) ?
The sum from n=0 to infinity of 1/(9-n^2) is converges