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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

d er i v a t i v e f (x) = \frac{( 3 x + 2 )}{( 3 x - 2 )}

\int (x ln (x)) d x

\frac{d}{d x} (ln (x y^{2}))

\frac{d}{d x} (\frac{1}{6 x ^{2}})

\int \frac{1}{( x ^{2} - 4 x + 4 ) ( x ^{2} - 5 x + 6 )} d x

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