What is the sum from n=1 to infinity of (8n)/(-2n+9) ?

The sum from n=1 to infinity of (8n)/(-2n+9) is diverges

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sum from n=1 to infinity of (8n)/(-2n+9)

n = 1 \sum \infty \frac{8 n}{- 2 n + 9}

diverges

Solution steps

n = 1 \sum \infty \frac{8 n}{- 2 n + 9}

Apply the constant multiplication rule:

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

= 8 \cdot n = 1 \sum \infty \frac{n}{- 2 n + 9}

Apply Series Divergence Test:

diverges

= 8 diverges

= diverges

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\int x^{2} x^{2} - 49 d x

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\int_{0.2}^{0.4} 200 x d x

What is the sum from n=1 to infinity of (8n)/(-2n+9) ?
The sum from n=1 to infinity of (8n)/(-2n+9) is diverges