What is the sum from k=5 to infinity of 1/((k-2)^4) ?

The sum from k=5 to infinity of 1/((k-2)^4) is converges

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sum from k=5 to infinity of 1/((k-2)^4)

k = 5 \sum \infty \frac{1}{( k - 2 ) ^{4}}

converges

Solution steps

k = 5 \sum \infty \frac{1}{( k - 2 ) ^{4}}

Apply Series Integral Test:

converges

= converges

\frac{d}{d x} (- \frac{2 e ^{\frac{2}{x}}}{x ^{2}})

f (x) = 2 tan (x)

\int - \frac{3 sin ( - x )}{cos ( - x )} d x

\int_{- 1}^{1} π (8 - 8 x^{2})^{2} d x

f (x) = x - (x + 1)^{4}

What is the sum from k=5 to infinity of 1/((k-2)^4) ?
The sum from k=5 to infinity of 1/((k-2)^4) is converges