What is the sum from n=0 to infinity of (5n)/(n^2+4) ?

The sum from n=0 to infinity of (5n)/(n^2+4) is diverges

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sum from n=0 to infinity of (5n)/(n^2+4)

n = 0 \sum \infty \frac{5 n}{n ^{2} + 4}

diverges

Solution steps

n = 0 \sum \infty \frac{5 n}{n ^{2} + 4}

Apply the constant multiplication rule:

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

= 5 \cdot n = 0 \sum \infty \frac{n}{n ^{2} + 4}

Apply Series Integral Test:

diverges

= 5 diverges

= diverges

f (x) = \frac{e ^{x}}{ln ( x )}

\int \frac{1}{( 2 + cos ( x ) )} d x

(t - 5)^{3}

\frac{d}{d x} (\frac{5 x ^{2}}{2})

(y) d x - 4 (x + y^{6}) d y = 0

What is the sum from n=0 to infinity of (5n)/(n^2+4) ?
The sum from n=0 to infinity of (5n)/(n^2+4) is diverges