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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= 2 converges

= converges

x \to 3 lim (\frac{4 x - 4}{5 x - 1})

\frac{d}{d x} (\frac{5 x}{x ^{2} + 7})

x \to 0 lim (\frac{x}{2 x})

\int_{0}^{\frac{π}{6}} 1 - cos (6 x) d x

x \to 0 lim ((1 + 7 x)^{\frac{1}{x}})

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