What is the sum from n=1 to infinity of 5/(n^2+6) ?

The sum from n=1 to infinity of 5/(n^2+6) is converges

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sum from n=1 to infinity of 5/(n^2+6)

n = 1 \sum \infty \frac{5}{n ^{2} + 6}

converges

Solution steps

n = 1 \sum \infty \frac{5}{n ^{2} + 6}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + 6}

Apply Series Comparison Test:

converges

= 5 converges

= converges

n = 100 \sum \infty \frac{1}{n}

n = 2 \sum \infty (- 0.4)^{n}

n = 0 \sum \infty π^{- 3 n}

n = 1 \sum \infty \frac{4 n}{n ^{2} + 1}

n = 0 \sum \infty 8 (\frac{2}{3})^{n}

What is the sum from n=1 to infinity of 5/(n^2+6) ?
The sum from n=1 to infinity of 5/(n^2+6) is converges