What is the sum from n=1 to infinity of n/(n^2+4) ?

The sum from n=1 to infinity of n/(n^2+4) is diverges

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sum from n=1 to infinity of n/(n^2+4)

n = 1 \sum \infty \frac{n}{n ^{2} + 4}

diverges

Solution steps

n = 1 \sum \infty \frac{n}{n ^{2} + 4}

Apply Series Comparison Test:

diverges

= diverges

\int 1 - y d y

\int 2 x ln (2 x) d x

y^{^{''}} - 6 y^{^{'}} + 10 y = 0

C (x) = \frac{4 x + 50}{x + 2}

\int x e^{11 x + 3} d x

What is the sum from n=1 to infinity of n/(n^2+4) ?
The sum from n=1 to infinity of n/(n^2+4) is diverges