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

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

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

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

\frac{d}{d x} (- 8 sin (\frac{3 x}{2} + 1))

x \to 4 - lim (2 x - 1)

f (x) = \frac{( x - 4 )}{x - 2}

\int_{3}^{5} 1

\int_{0}^{1.5} 1250 x d x

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