What is the sum from n=1 to infinity of 6/(n^2+n^3) ?

The sum from n=1 to infinity of 6/(n^2+n^3) is converges

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sum from n=1 to infinity of 6/(n^2+n^3)

n = 1 \sum \infty \frac{6}{n ^{2} + n ^{3}}

converges

Solution steps

n = 1 \sum \infty \frac{6}{n ^{2} + n ^{3}}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + n ^{3}}

Apply Series Comparison Test:

converges

= 6 converges

= converges

n = 1 \sum \infty \frac{9}{n ^{2} + 4}

n = 1 \sum \infty \frac{n ^{5}}{8 ^{n}}

n = 0 \sum \infty \frac{n + 1}{2 n + 3}

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

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

What is the sum from n=1 to infinity of 6/(n^2+n^3) ?
The sum from n=1 to infinity of 6/(n^2+n^3) is converges