What is the sum from n=0 to infinity of n/(n^2+4) ?

The sum from n=0 to infinity of n/(n^2+4) is diverges

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sum from n=0 to infinity of n/(n^2+4)

n = 0 \sum \infty \frac{n}{n ^{2} + 4}

diverges

Solution steps

n = 0 \sum \infty \frac{n}{n ^{2} + 4}

Apply Series Integral Test:

diverges

= diverges

\int_{0}^{1} 9

\int 4 x^{9} e^{- x^{5}} d x

y = 2 x, y = 5 x^{2}

\int (- 4 x) d x

x \to - 6 + lim (\frac{x + 5}{x + 6})

What is the sum from n=0 to infinity of n/(n^2+4) ?
The sum from n=0 to infinity of n/(n^2+4) is diverges