What is the sum from n=0 to infinity of n/(9+n^2) ?

The sum from n=0 to infinity of n/(9+n^2) is diverges

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sum from n=0 to infinity of n/(9+n^2)

n = 0 \sum \infty \frac{n}{9 + n ^{2}}

diverges

Solution steps

n = 0 \sum \infty \frac{n}{9 + n ^{2}}

Apply Series Integral Test:

diverges

= diverges

n = 0 \sum \infty 7 (\frac{2}{3})^{n + 2}

n = 1 \sum \infty \frac{e ^{\frac{8}{n}}}{n}

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

n = 0 \sum \infty e^{- 7 n}

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

What is the sum from n=0 to infinity of n/(9+n^2) ?
The sum from n=0 to infinity of n/(9+n^2) is diverges