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

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

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

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

converges

Solution steps

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

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

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

Apply Series Integral Test:

converges

= converges

n = 0 \sum \infty \frac{3}{2 n}

n = 0 \sum \infty \frac{1}{6} (\frac{5}{6})^{n}

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

n = 1 \sum \infty \frac{10 0 ^{n}}{n !}

n = 1 \sum \infty n ln (n)

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