What is the sum from n=6 to infinity of 1/((n-3)^4) ?

The sum from n=6 to infinity of 1/((n-3)^4) is converges

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sum from n=6 to infinity of 1/((n-3)^4)

n = 6 \sum \infty \frac{1}{( n - 3 ) ^{4}}

converges

Solution steps

n = 6 \sum \infty \frac{1}{( n - 3 ) ^{4}}

Apply Series Integral Test:

converges

= converges

n = 1 \sum \infty \frac{1}{n ^{7} + 8}

n = 0 \sum \infty \frac{3}{- 3 ^{n}}

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

k = 1 \sum \infty \frac{6}{k}

n = 1 \sum \infty \frac{1}{3 n ^{2} + 9}

What is the sum from n=6 to infinity of 1/((n-3)^4) ?
The sum from n=6 to infinity of 1/((n-3)^4) is converges