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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

converges

= 3 converges

= converges

\frac{d}{d x} (\frac{1 - x}{x})

{\frac{s + 1}{s ^{2} + 2}}

\int_{1}^{3} ln (x - 1) d x

n = 1 \sum \infty cos (2 n)

y = x^{2} - 1, y = x + 2, x = 0, x + 1

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