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

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

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

n = 1 \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

n = 0 \sum \infty cos (\frac{nπ}{3})

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

n = 1 \sum \infty - 8 (\frac{4}{5})^{n - 1}

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

n = 7 \sum \infty 2^{- n}

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