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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

converges

= 2 converges

= converges

n = 1 \sum \infty \frac{9 ^{n}}{n ^{5}}

n = 1 \sum \infty \frac{1}{9 n ^{3}}

n = 0 \sum \infty 5 (\frac{7}{8})^{n}

n = 0 \sum \infty \frac{1}{n ^{6} + 1}

n = 0 \sum \infty \frac{3 n}{n !}

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