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

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

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

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

converges

Solution steps

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

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} + 3}

Apply Series Comparison Test:

converges

= 3 converges

= converges

n = 0 \sum \infty n^{^{'}} + n

n = 0 \sum \infty n^{^{'}} - n

n = 1 \sum \infty n! (- e)^{- n}

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

n = 0 \sum \infty 4 (\frac{2}{3})^{n}

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