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

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

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

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

converges

Solution steps

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

Apply Series Integral Test:

converges

= converges

k = 1 \sum \infty \frac{ln ( k )}{k}

n = 1 \sum \infty - 9 (\frac{1}{6})^{n - 1}

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

k = 1 \sum \infty \frac{1}{2 k ^{2} + k}

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

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