What is the sum from n=1 to infinity of 7/(n^2+81) ?

The sum from n=1 to infinity of 7/(n^2+81) 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 7/(n^2+81)

n = 1 \sum \infty \frac{7}{n ^{2} + 81}

converges

Solution steps

n = 1 \sum \infty \frac{7}{n ^{2} + 81}

Apply the constant multiplication rule:

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

= 7 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + 81}

Apply Series Comparison Test:

converges

= 7 converges

= converges

f (x) = - x^{2} + 25

\int (x + 1)^{2} d x

\int cosh (m t) d t

\int_{- 2}^{5} 6 d x

y^{^{''}} - 3 y^{^{'}} = 9 t

What is the sum from n=1 to infinity of 7/(n^2+81) ?
The sum from n=1 to infinity of 7/(n^2+81) is converges