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

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

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

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

converges

Solution steps

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

Apply Series Integral Test:

converges

= converges

\frac{d}{d x} (4 x^{5} - 5 x^{4})

x^{3} + y^{3} - 6 x y = 0, (\frac{4}{3}, \frac{8}{3})

f (x) = \frac{x ^{2} - 1}{x ^{2} + 1}, at x = 0

\frac{\partial}{\partial x} (- 4 cos (2 x))

\frac{d ^{2} y}{d x ^{2}} - 4 y = (x^{2} - 3) sin (2 x)

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