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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Series Comparison Test:

converges

= 2 converges

= converges

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

f (x) = 6 x + 10 - 19 e^{x}, (0, - 9)

\int \frac{9 x e ^{2 x}}{( 1 + 2 x ) ^{2}} d x

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