What is the sum from n=1 to infinity of 9/(n^2+49) ?

The sum from n=1 to infinity of 9/(n^2+49) is converges

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sum from n=1 to infinity of 9/(n^2+49)

n = 1 \sum \infty \frac{9}{n ^{2} + 49}

converges

Solution steps

n = 1 \sum \infty \frac{9}{n ^{2} + 49}

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} + 49}

Apply Series Comparison Test:

converges

= 9 converges

= converges

n = 0 \sum \infty 3 (\frac{4}{5})^{n}

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

k = 1 \sum \infty \frac{k !}{( 2 k ) !}

n = 1 \sum \infty (\frac{- e}{π})^{n}

n = 0 \sum \infty 3 (- 1)^{n}

What is the sum from n=1 to infinity of 9/(n^2+49) ?
The sum from n=1 to infinity of 9/(n^2+49) is converges