What is the sum from n=1 to infinity of 5/(n^3+n) ?

The sum from n=1 to infinity of 5/(n^3+n) is converges

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sum from n=1 to infinity of 5/(n^3+n)

n = 1 \sum \infty \frac{5}{n ^{3} + n}

converges

Solution steps

n = 1 \sum \infty \frac{5}{n ^{3} + n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 5 \cdot n = 1 \sum \infty \frac{1}{n ^{3} + n}

Apply Series Comparison Test:

converges

= 5 converges

= converges

n = 1 \sum \infty (- e)^{- n}

n = 0 \sum \infty (- \frac{5}{7})^{n}

n = 1 \sum \infty \frac{9}{n}

k = 2 \sum \infty \frac{1}{k ln ( k )}

n = 1 \sum \infty cos (\frac{π}{n})

What is the sum from n=1 to infinity of 5/(n^3+n) ?
The sum from n=1 to infinity of 5/(n^3+n) is converges