What is the sum from n=1 to infinity of 3/(n^2+5) ?

The sum from n=1 to infinity of 3/(n^2+5) is converges

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sum from n=1 to infinity of 3/(n^2+5)

n = 1 \sum \infty \frac{3}{n ^{2} + 5}

converges

Solution steps

n = 1 \sum \infty \frac{3}{n ^{2} + 5}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + 5}

Apply Series Comparison Test:

converges

= 3 converges

= converges

\int sec (6 x) d x

\int x^{2} e^{- 10 x} d x

\int - \frac{1}{x ^{3}} d x

x \to 5 + lim (I n (x^{2} - 25))

y^{^{''}} + 2 y^{^{'}} + y = t^{2} + 1 - e^{t}

What is the sum from n=1 to infinity of 3/(n^2+5) ?
The sum from n=1 to infinity of 3/(n^2+5) is converges