What is the sum from n=1 to infinity of 9/(n^2+64) ?

The sum from n=1 to infinity of 9/(n^2+64) 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 9/(n^2+64)

n = 1 \sum \infty \frac{9}{n ^{2} + 64}

converges

Solution steps

n = 1 \sum \infty \frac{9}{n ^{2} + 64}

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + 64}

Apply Series Comparison Test:

converges

= 9 converges

= converges

n = 6 \sum \infty \frac{1}{n} - \frac{1}{n + 3}

n = 0 \sum \infty \frac{1}{8 n - 1}

n = 1 \sum \infty (sin (200))^{n}

n = 0 \sum \infty \frac{3 n}{6 + n ^{2}}

n = 1 \sum \infty \frac{4}{3 ^{n} + 8}

What is the sum from n=1 to infinity of 9/(n^2+64) ?
The sum from n=1 to infinity of 9/(n^2+64) is converges