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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

n = 0 \sum \infty n^{2} e^{n}

n = 0 \sum \infty \frac{n ^{3}}{2 - n}

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

n = 1 \sum \infty \frac{6}{n} - \frac{6}{n ^{2}}

n = 1 \sum \infty 30 (0.1)^{n}

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