What is the sum from k=6 to infinity of 1/((k-5)^4) ?

The sum from k=6 to infinity of 1/((k-5)^4) is converges

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sum from k=6 to infinity of 1/((k-5)^4)

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

converges

Solution steps

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

Apply Series Integral Test:

converges

= converges

n = 1 \sum \infty 53 \cdot (\frac{1}{2})^{n}

n = 0 \sum \infty \frac{1}{ln ( 6 n )}

n = 1 \sum \infty \frac{1}{12 n + 1}

k = 0 \sum \infty \frac{1}{k}

n = 1 \sum \infty (- \frac{6}{25})^{n}

What is the sum from k=6 to infinity of 1/((k-5)^4) ?
The sum from k=6 to infinity of 1/((k-5)^4) is converges