What is the sum from n=0 to infinity of n(n-6) ?

The sum from n=0 to infinity of n(n-6) is diverges

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sum from n=0 to infinity of n(n-6)

n = 0 \sum \infty n (n - 6)

diverges

Solution steps

n = 0 \sum \infty n (n - 6)

Expand

n (n - 6) : n^{2} - 6 n

= n = 0 \sum \infty n^{2} - 6 n

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty 6 (\frac{1}{8})^{n - 1}

n = 0 \sum \infty 3 n ln (1 + n)

n = 1 \sum \infty n^{4} + 1

n = 1 \sum \infty (\frac{5}{6})^{4 n - 5}

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

What is the sum from n=0 to infinity of n(n-6) ?
The sum from n=0 to infinity of n(n-6) is diverges