What is the sum from n=1 to infinity of 4/(3^n+8) ?

The sum from n=1 to infinity of 4/(3^n+8) is converges

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sum from n=1 to infinity of 4/(3^n+8)

n = 1 \sum \infty \frac{4}{3 ^{n} + 8}

converges

Solution steps

n = 1 \sum \infty \frac{4}{3 ^{n} + 8}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty \frac{1}{3 ^{n} + 8}

Apply Series Ratio Test:

converges

= 4 converges

= converges

k = 1 \sum \infty (cos (17))^{k}

n = 1 \sum \infty \frac{8}{1 0 ^{n}}

n = 1 \sum \infty cos (\frac{15}{n})

n = 0 \sum \infty \frac{i}{2 ^{n}}

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

What is the sum from n=1 to infinity of 4/(3^n+8) ?
The sum from n=1 to infinity of 4/(3^n+8) is converges