What is the sum from n=1 to infinity of (3n)/(n^2-7) ?

The sum from n=1 to infinity of (3n)/(n^2-7) is diverges

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sum from n=1 to infinity of (3n)/(n^2-7)

n = 1 \sum \infty \frac{3 n}{n ^{2} - 7}

diverges

Solution steps

n = 1 \sum \infty \frac{3 n}{n ^{2} - 7}

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} - 7}

Apply Series Integral Test:

diverges

= 3 diverges

= diverges

n = 1 \sum \infty 2 e^{- 2 n}

n = 0 \sum \infty \frac{1}{1 - 4 n}

n = 1 \sum \infty \frac{9}{n ^{2} + 3 n}

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

n = 0 \sum \infty \frac{1}{9 - n}

What is the sum from n=1 to infinity of (3n)/(n^2-7) ?
The sum from n=1 to infinity of (3n)/(n^2-7) is diverges