What is the sum from n=1 to infinity of 5/(n^2+4) ?

The sum from n=1 to infinity of 5/(n^2+4) is converges

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sum from n=1 to infinity of 5/(n^2+4)

n = 1 \sum \infty \frac{5}{n ^{2} + 4}

converges

Solution steps

n = 1 \sum \infty \frac{5}{n ^{2} + 4}

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} + 4}

Apply Series Comparison Test:

converges

= 5 converges

= converges

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

j = 1 \sum \infty 3^{- 3 j}

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

n = 0 \sum \infty 7^{n}

n = 1 \sum \infty \frac{( - \frac{1}{5} ) ^{n}}{n}

What is the sum from n=1 to infinity of 5/(n^2+4) ?
The sum from n=1 to infinity of 5/(n^2+4) is converges