What is the sum from n=0 to infinity of 7/(4-n^2) ?

The sum from n=0 to infinity of 7/(4-n^2) is converges

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sum from n=0 to infinity of 7/(4-n^2)

n = 0 \sum \infty \frac{7}{4 - n ^{2}}

converges

Solution steps

n = 3 \sum \infty \frac{7}{4 - n ^{2}}

Apply the constant multiplication rule:

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

= 7 \cdot n = 3 \sum \infty \frac{1}{4 - n ^{2}}

Factor

4 - n^{2} : - (- 4 + n^{2})

= 7 \cdot n = 3 \sum \infty \frac{1}{- ( - 4 + n ^{2} )}

Apply the constant multiplication rule:

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

= 7 (- n = 3 \sum \infty \frac{1}{- 4 + n ^{2}})

Apply Series Integral Test:

converges

= 7 (- converges)

= converges

n = 1 \sum \infty n (\frac{1}{4})^{n}

k = 1 \sum \infty \frac{5}{k + 6}

n = 1 \sum \infty \frac{1}{n} - \frac{1}{n + 5}

n = 1 \sum \infty (\frac{8}{5})^{n}

n = 0 \sum \infty 20 (\frac{1}{4})^{n}

What is the sum from n=0 to infinity of 7/(4-n^2) ?
The sum from n=0 to infinity of 7/(4-n^2) is converges