What is the sum from n=1 to infinity of (3n)/(n^2+4) ?

The sum from n=1 to infinity of (3n)/(n^2+4) is diverges

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sum from n=1 to infinity of (3n)/(n^2+4)

n = 1 \sum \infty \frac{3 n}{n ^{2} + 4}

diverges

Solution steps

n = 1 \sum \infty \frac{3 n}{n ^{2} + 4}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{n}{n ^{2} + 4}

Apply Series Comparison Test:

diverges

= 3 diverges

= diverges

\int cos^{2} (k x) d x

\int_{2}^{4} \frac{x}{x ^{2} + 6 x + 13} d x

n = 0 \sum \infty \frac{1}{n ^{2}}

\frac{\partial}{\partial x} (e^{6 x + 9 y})

y = \frac{1}{2} x, y = 1, y = \frac{1}{25} x^{2}

What is the sum from n=1 to infinity of (3n)/(n^2+4) ?
The sum from n=1 to infinity of (3n)/(n^2+4) is diverges