What is the sum from n=0 to infinity of 7/(n^2+9) ?

The sum from n=0 to infinity of 7/(n^2+9) 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=0 to infinity of 7/(n^2+9)

n = 0 \sum \infty \frac{7}{n ^{2} + 9}

converges

Solution steps

n = 0 \sum \infty \frac{7}{n ^{2} + 9}

Apply the constant multiplication rule:

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

= 7 \cdot n = 0 \sum \infty \frac{1}{n ^{2} + 9}

Apply Series Comparison Test:

converges

= 7 converges

= converges

n = 1 \sum \infty \frac{9}{n ^{2} + n ^{3}}

k = 3 \sum \infty \frac{7}{k ln ^{2} ( k )}

k = 2 \sum \infty \frac{k}{ln ( 5 k )}

n = 2 \sum \infty (\frac{1}{9 n - 2})^{7}

x = 0 \sum \infty x (x)

What is the sum from n=0 to infinity of 7/(n^2+9) ?
The sum from n=0 to infinity of 7/(n^2+9) is converges