What is the sum from n=1 to infinity of 1/(n^3-5n^2) ?

The sum from n=1 to infinity of 1/(n^3-5n^2) is converges

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sum from n=1 to infinity of 1/(n^3-5n^2)

n = 1 \sum \infty \frac{1}{n ^{3} - 5 n ^{2}}

converges

Solution steps

n = 6 \sum \infty \frac{1}{n ^{3} - 5 n ^{2}}

Apply Series Integral Test:

converges

= converges

k = 1 \sum \infty \frac{k ^{2}}{4 ^{k}}

n = 0 \sum \infty n 2^{n}

n = 1 \sum \infty \frac{1}{( 2 ^{n} )}

n = 1 \sum \infty 2^{n}

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

What is the sum from n=1 to infinity of 1/(n^3-5n^2) ?
The sum from n=1 to infinity of 1/(n^3-5n^2) is converges