What is the sum from n=1 to infinity of 5/(n^2+121) ?

The sum from n=1 to infinity of 5/(n^2+121) 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+121)

n = 1 \sum \infty \frac{5}{n ^{2} + 121}

converges

Solution steps

n = 1 \sum \infty \frac{5}{n ^{2} + 121}

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} + 121}

Apply Series Comparison Test:

converges

= 5 converges

= converges

n = 2 \sum \infty \frac{7 n}{ln ( n )}

n = 1 \sum \infty \frac{6 n ^{2}}{n !}

n = 1 \sum \infty - 7 (\frac{3}{8})^{n}

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

n = 0 \sum \infty 2^{\frac{n}{2}}

What is the sum from n=1 to infinity of 5/(n^2+121) ?
The sum from n=1 to infinity of 5/(n^2+121) is converges