What is the sum from n=1 to infinity of 7/(n^2+36) ?

The sum from n=1 to infinity of 7/(n^2+36) is converges

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sum from n=1 to infinity of 7/(n^2+36)

n = 1 \sum \infty \frac{7}{n ^{2} + 36}

converges

Solution steps

n = 1 \sum \infty \frac{7}{n ^{2} + 36}

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} + 36}

Apply Series Comparison Test:

converges

= 7 converges

= converges

x \to 0 lim (∣ x ∣)

\int_{0}^{π} ∣ cos (x) ∣ d x

x \to 0 lim (- \frac{1}{1} + \frac{1}{x})

arccos (x^{2} + \frac{3}{2} x)

\frac{d}{d x} (e^{- 8 x} cos (8 x))

What is the sum from n=1 to infinity of 7/(n^2+36) ?
The sum from n=1 to infinity of 7/(n^2+36) is converges