What is the sum from n=1 to infinity of 1/(n^2+9) ?

The sum from n=1 to infinity of 1/(n^2+9) is converges

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sum from n=1 to infinity of 1/(n^2+9)

n = 1 \sum \infty \frac{1}{n ^{2} + 9}

converges

Solution steps

n = 1 \sum \infty \frac{1}{n ^{2} + 9}

Apply Series Comparison Test:

converges

= converges

\frac{d}{d x} (- \frac{18}{x ^{3}})

f (x) = \frac{1}{x - 2}, at x = 4

(\frac{19}{2}, - \frac{5}{3}), (- \frac{3}{2}, - \frac{7}{24})

\int \frac{sec ( x ) tan ( x )}{sec ( x ) - 4} d x

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = 4

What is the sum from n=1 to infinity of 1/(n^2+9) ?
The sum from n=1 to infinity of 1/(n^2+9) is converges