What is the sum from n=1 to infinity of 5/(n^2+36) ?

The sum from n=1 to infinity of 5/(n^2+36) 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 5/(n^2+36)

n = 1 \sum \infty \frac{5}{n ^{2} + 36}

converges

Solution steps

n = 1 \sum \infty \frac{5}{n ^{2} + 36}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + 36}

Apply Series Comparison Test:

converges

= 5 converges

= converges

n = 2 \sum \infty \frac{1}{ln ( n ^{2} )}

k = 1 \sum \infty (cos (10))^{k}

n = 1 \sum \infty \frac{( - 1 ) ^{n}}{3}

n = 1 \sum \infty cos (\frac{16}{n})

n = 1 \sum \infty \frac{1}{n ^{4} + 2 n + 8}

What is the sum from n=1 to infinity of 5/(n^2+36) ?
The sum from n=1 to infinity of 5/(n^2+36) is converges