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

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

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

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

converges

Solution steps

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

Take the partial fraction of

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

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

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

converges

= - converges

= converges

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

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

x = 1 \sum \infty \frac{1}{x ( x + 1 )}

n = 1 \sum \infty \frac{1}{( 2 n + 1 ) !}

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

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