What is the sum from n=1 to infinity of 9/(n^2+16) ?

The sum from n=1 to infinity of 9/(n^2+16) is converges

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sum from n=1 to infinity of 9/(n^2+16)

n = 1 \sum \infty \frac{9}{n ^{2} + 16}

converges

Solution steps

n = 1 \sum \infty \frac{9}{n ^{2} + 16}

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + 16}

Apply Series Comparison Test:

converges

= 9 converges

= converges

n = 0 \sum \infty \frac{1}{ln ( n ^{n} )}

n = 0 \sum \infty \frac{1}{4 + n ^{2}}

n = 1 \sum \infty \frac{ln ( 7 n )}{n}

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

n = 0 \sum \infty 2^{n} (\frac{1}{2})^{n}

What is the sum from n=1 to infinity of 9/(n^2+16) ?
The sum from n=1 to infinity of 9/(n^2+16) is converges