What is the sum from n=0 to infinity of 4/(9-n^2) ?

The sum from n=0 to infinity of 4/(9-n^2) is converges

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sum from n=0 to infinity of 4/(9-n^2)

n = 0 \sum \infty \frac{4}{9 - n ^{2}}

converges

Solution steps

n = 4 \sum \infty \frac{4}{9 - n ^{2}}

Apply the constant multiplication rule:

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

= 4 \cdot n = 4 \sum \infty \frac{1}{9 - n ^{2}}

Factor

9 - n^{2} : - (- 9 + n^{2})

= 4 \cdot n = 4 \sum \infty \frac{1}{- ( - 9 + n ^{2} )}

Apply the constant multiplication rule:

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

= 4 (- n = 4 \sum \infty \frac{1}{- 9 + n ^{2}})

Apply Series Integral Test:

converges

= 4 (- converges)

= converges

k = 1 \sum \infty \frac{3}{k} - \frac{3}{k + 2}

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

n = 1 \sum \infty \frac{1}{n} - \frac{2}{3 n - 1}

k = 2 \sum \infty (- 0.95)^{k}

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

What is the sum from n=0 to infinity of 4/(9-n^2) ?
The sum from n=0 to infinity of 4/(9-n^2) is converges