What is the sum from k=1 to infinity of 8/(k^7+9) ?

The sum from k=1 to infinity of 8/(k^7+9) is converges

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sum from k=1 to infinity of 8/(k^7+9)

k = 1 \sum \infty \frac{8}{k ^{7} + 9}

converges

Solution steps

k = 1 \sum \infty \frac{8}{k ^{7} + 9}

Apply the constant multiplication rule:

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

= 8 \cdot k = 1 \sum \infty \frac{1}{k ^{7} + 9}

Apply Series Comparison Test:

converges

= 8 converges

= converges

n = 0 \sum \infty \frac{2 n}{5 n + 1}

n = 1 \sum \infty n e^{- \frac{n}{9}}

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

n = 1 \sum \infty \frac{6}{4 n ^{2} + 4 n}

k = 9 \sum \infty \frac{7}{k ^{6.6}}

What is the sum from k=1 to infinity of 8/(k^7+9) ?
The sum from k=1 to infinity of 8/(k^7+9) is converges