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}}

Apply Series Comparison Test:

converges

= converges

v (t) = \frac{4 t ^{2} + 2}{5 t + 5}

\int (x + 2) 2 - x d x

y^{^{'}} = k y (1 - y)

\frac{d}{d x} (\frac{1}{x} + ln (x))

5 x^{2} - x^{3}

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