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

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

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

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

converges

Solution steps

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

Apply Series Integral Test:

converges

= converges

f (x) = - x^{3} + 2 x^{2} - 4, at x = - 1

\frac{d}{d x} (x^{2} - 2 x y + y^{2})

(3 x^{7} - 4) (2 x - 5 x^{2})

x \to - 2 lim (1 + x)

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