What is the sum from n=0 to infinity of 1/((2+2n)^3) ?

The sum from n=0 to infinity of 1/((2+2n)^3) is converges

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sum from n=0 to infinity of 1/((2+2n)^3)

n = 0 \sum \infty \frac{1}{( 2 + 2 n ) ^{3}}

converges

Solution steps

n = 0 \sum \infty \frac{1}{( 2 + 2 n ) ^{3}}

Take the partial fraction of

\frac{1}{( 2 + 2 n ) ^{3}} : \frac{1}{8 ( n + 1 ) ^{3}}

= n = 0 \sum \infty \frac{1}{8 ( n + 1 ) ^{3}}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= \frac{1}{8} \cdot n = 0 \sum \infty \frac{1}{( n + 1 ) ^{3}}

Apply Series Comparison Test:

converges

= \frac{1}{8} converges

= converges

\int_{- 1}^{3} 2 x^{2} + 8 d x

\int 4 y^{5} d y

x (x - 8)

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

\frac{\partial}{\partial y} (e^{- y} (x^{2} - y^{2}))

What is the sum from n=0 to infinity of 1/((2+2n)^3) ?
The sum from n=0 to infinity of 1/((2+2n)^3) is converges