What is the sum from n=0 to infinity of 3/(3+n^2) ?

The sum from n=0 to infinity of 3/(3+n^2) is converges

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sum from n=0 to infinity of 3/(3+n^2)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

= 3 \cdot n = 0 \sum \infty \frac{1}{3 + n ^{2}}

Apply Series Comparison Test:

converges

= 3 converges

= converges

n = 1 \sum \infty \frac{n}{n ^{2} + 8}

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

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

n = 3 \sum \infty n e^{- n}

n = 1 \sum \infty \frac{5}{n ^{2} + 49}

What is the sum from n=0 to infinity of 3/(3+n^2) ?
The sum from n=0 to infinity of 3/(3+n^2) is converges