What is the sum from k=6 to infinity of 1/((k-5)^3) ?

The sum from k=6 to infinity of 1/((k-5)^3) is converges

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sum from k=6 to infinity of 1/((k-5)^3)

k = 6 \sum \infty \frac{1}{( k - 5 ) ^{3}}

converges

Solution steps

k = 6 \sum \infty \frac{1}{( k - 5 ) ^{3}}

Apply Series Integral Test:

converges

= converges

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

n = 1 \sum \infty \frac{1}{n ^{2} + 7 n + 1}

k = 0 \sum \infty (- \frac{1}{2})^{k}

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

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

What is the sum from k=6 to infinity of 1/((k-5)^3) ?
The sum from k=6 to infinity of 1/((k-5)^3) is converges