What is the sum from n=1 to infinity of n/(n^2+6) ?

The sum from n=1 to infinity of n/(n^2+6) is diverges

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sum from n=1 to infinity of n/(n^2+6)

n = 1 \sum \infty \frac{n}{n ^{2} + 6}

diverges

Solution steps

n = 1 \sum \infty \frac{n}{n ^{2} + 6}

Apply Series Comparison Test:

diverges

= diverges

k = 1 \sum \infty \frac{5}{k} - \frac{5}{k + 1}

n = 1 \sum \infty \frac{5}{n ( n + 1 )}

k = 1 \sum \infty \frac{4}{k}

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

n = 1 \sum \infty \frac{5}{n ^{2} + 36}

What is the sum from n=1 to infinity of n/(n^2+6) ?
The sum from n=1 to infinity of n/(n^2+6) is diverges