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

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

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sum from n=2 to infinity of (3n)/(n^2-2)

n = 2 \sum \infty \frac{3 n}{n ^{2} - 2}

diverges

Solution steps

n = 2 \sum \infty \frac{3 n}{n ^{2} - 2}

Apply the constant multiplication rule:

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

= 3 \cdot n = 2 \sum \infty \frac{n}{n ^{2} - 2}

Apply Series Comparison Test:

diverges

= 3 diverges

= diverges

x \to 5 + lim (\frac{6 x}{x - 5})

\int 3 x 5 x^{2} + 1 d x

\frac{d y}{d x} = (x + 2)^{2}

\frac{d}{d x} (arctan (\frac{y}{x}))

4 t \frac{d y}{d t} + y = t^{6}

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