What is the sum from k=3 to infinity of 1/((k-2)^5) ?

The sum from k=3 to infinity of 1/((k-2)^5) is converges

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sum from k=3 to infinity of 1/((k-2)^5)

k = 3 \sum \infty \frac{1}{( k - 2 ) ^{5}}

converges

Solution steps

k = 3 \sum \infty \frac{1}{( k - 2 ) ^{5}}

Apply Series Integral Test:

converges

= converges

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

n = 1 \sum \infty \frac{n ^{4}}{5 ^{n}}

n = 0 \sum \infty \frac{n + 1}{2 n + 1}

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

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

What is the sum from k=3 to infinity of 1/((k-2)^5) ?
The sum from k=3 to infinity of 1/((k-2)^5) is converges