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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply exponent rule:

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

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

Apply Series Integral Test:

converges

= 3 converges

= converges

n = 0 \sum \infty \frac{1}{n}

n = 0 \sum \infty (\frac{4}{11})^{n}

j = 1 \sum \infty 2 (- \frac{3}{5})^{j}

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

n = 1 \sum \infty 4 (\frac{π}{4})^{n}

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