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 diverges

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

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

diverges

Solution steps

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

\frac{1}{n} = \frac{1 + n}{n} - 1

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

Apply Series Divergence Test:

diverges

= diverges

\frac{1}{s ^{2} - 3}

\int \frac{2 x}{( 1 - x ^{2} )} d x

y^{^{'}} = \frac{y}{x} + x e^{x}

\int \frac{2}{x ^{2} ( x ^{2} + 25 )} d x

\frac{\partial}{\partial y} (6 x y - 6 y)

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 diverges