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

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

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

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

converges

Solution steps

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

(n - 1)^{2} = n^{2} - 2 n + 1

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

Apply Series Comparison Test:

converges

= converges

\frac{x}{e ^{x}}

\frac{d y}{d x} = e^{5 x + 6 y}

1 + e^{2 x}

\int sec^{2} (x) - 5 d x

\frac{d y}{d t} = 2 \cdot y

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