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

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

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

n = 1 \sum \infty \frac{n}{n + 8}

diverges

Solution steps

n = 1 \sum \infty \frac{n}{n + 8}

Apply Series Divergence Test:

diverges

= diverges

n = 9 \sum \infty \frac{1}{9 n ^{4} + 8}

n = 0 \sum \infty (0. 2^{n}) 0.7

n = 0 \sum \infty (0. 2^{n}) 0.8

k = 7 \sum \infty \frac{1}{k ^{2} - 5 k + 6}

k = 0 \sum \infty (\frac{4}{7})^{k}

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