What is the sum from n=1 to infinity of 5/(n^2+9) ?

The sum from n=1 to infinity of 5/(n^2+9) is converges

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sum from n=1 to infinity of 5/(n^2+9)

n = 1 \sum \infty \frac{5}{n ^{2} + 9}

converges

Solution steps

n = 1 \sum \infty \frac{5}{n ^{2} + 9}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + 9}

Apply Series Comparison Test:

converges

= 5 converges

= converges

\int_{\frac{1}{5}}^{3} 6 x ln (5 x) d x

\frac{2 x}{x - 1}

y^{^{'}} = \frac{1}{( t - 1 ) ( y + 1 )}, y (0) = 1

\int \frac{1}{s ^{2} - 2 s + 5} d s

x \to - 4 lim (\frac{4 - ∣ x ∣}{4 + x})

What is the sum from n=1 to infinity of 5/(n^2+9) ?
The sum from n=1 to infinity of 5/(n^2+9) is converges