What is the sum from n=5 to infinity of 1/((n-4)^3) ?

The sum from n=5 to infinity of 1/((n-4)^3) is converges

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sum from n=5 to infinity of 1/((n-4)^3)

n = 5 \sum \infty \frac{1}{( n - 4 ) ^{3}}

converges

Solution steps

n = 5 \sum \infty \frac{1}{( n - 4 ) ^{3}}

Apply Series Integral Test:

converges

= converges

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

k = 5 \sum \infty \frac{6}{k ln ^{2} ( k )}

n = 0 \sum \infty \frac{i}{n ^{4}}

n = 0 \sum \infty \frac{n}{11 n ^{6} + 1}

n = 0 \sum \infty (0.25)^{n}

What is the sum from n=5 to infinity of 1/((n-4)^3) ?
The sum from n=5 to infinity of 1/((n-4)^3) is converges