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

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

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

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

converges

Solution steps

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

Expand

n^{2} (n + 1) : n^{3} + n^{2}

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

Apply Series Comparison Test:

converges

= converges

n = 1 \sum \infty \frac{n ^{4}}{4 n !}

x = 1 \sum \infty \frac{x}{x ( x + 1 )}

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

k = 1 \sum \infty k e^{- 16 k}

k = 1 \sum \infty k e^{- 15 k}

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