What is the sum from n=1 to infinity of 1/((3+2n)^2) ?

The sum from n=1 to infinity of 1/((3+2n)^2) is converges

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sum from n=1 to infinity of 1/((3+2n)^2)

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

\int_{1}^{\infty} 21 \frac{e - x}{x} d x

\int - 3 ln (2 x + 1) d x

7 e^{x} arctan (x)

t^{2} y^{^{''}} - 2 t y^{^{'}} - 4 y = 0

What is the sum from n=1 to infinity of 1/((3+2n)^2) ?
The sum from n=1 to infinity of 1/((3+2n)^2) is converges