What is the sum from n=1 to infinity of n/(n^3+2) ?

The sum from n=1 to infinity of n/(n^3+2) is converges

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sum from n=1 to infinity of n/(n^3+2)

n = 1 \sum \infty \frac{n}{n ^{3} + 2}

converges

Solution steps

n = 1 \sum \infty \frac{n}{n ^{3} + 2}

Apply Series Comparison Test:

converges

= converges

n = 0 \sum \infty \frac{1}{2 ^{n + 2}}

\int \frac{x ^{2}}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

\frac{d y}{d x} = - \frac{2}{x}

\int \frac{x}{( x ^{2} + 4 ) ^{3}} d x

\int \frac{3}{x ^{2} 1 + x ^{2}} d x

What is the sum from n=1 to infinity of n/(n^3+2) ?
The sum from n=1 to infinity of n/(n^3+2) is converges