What is the sum from n=1 to infinity of 7/(n^2+16) ?

The sum from n=1 to infinity of 7/(n^2+16) is converges

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sum from n=1 to infinity of 7/(n^2+16)

n = 1 \sum \infty \frac{7}{n ^{2} + 16}

converges

Solution steps

n = 1 \sum \infty \frac{7}{n ^{2} + 16}

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} + 16}

Apply Series Comparison Test:

converges

= 7 converges

= converges

n = 0 \sum \infty \frac{n}{7 n ^{3} + 1}

n = 0 \sum \infty ln (2 - n)

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

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

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

What is the sum from n=1 to infinity of 7/(n^2+16) ?
The sum from n=1 to infinity of 7/(n^2+16) is converges