What is the sum from n=0 to infinity of 1/((1+n)^2) ?

The sum from n=0 to infinity of 1/((1+n)^2) is converges

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sum from n=0 to infinity of 1/((1+n)^2)

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

\frac{d}{d x} (3 cos (3 x))

4 x^{\frac{5}{4}} + 10 x^{\frac{3}{2}} + 5 x

(1, - 2), (3, 1)

y^{^{''}} + 36 y = - 60 sin (6 t)

\frac{d y}{d t} = - 2 (1 + y^{2}) tan (2 t)

What is the sum from n=0 to infinity of 1/((1+n)^2) ?
The sum from n=0 to infinity of 1/((1+n)^2) is converges