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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Take the partial fraction of

\frac{1}{( 1 - n ) ^{3}} : - \frac{1}{( n - 1 ) ^{3}}

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

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

converges

= 2 (- converges)

= converges

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

n = 1 \sum \infty \frac{n !}{11 4 ^{n}}

n = 7 \sum \infty \frac{1}{n} - \frac{1}{n + 3}

n = 1 \sum \infty n + 1

n = 0 \sum \infty (1 + \frac{5}{n})^{n}

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