What is the sum from n=1 to infinity of 9/(6+9^n) ?

The sum from n=1 to infinity of 9/(6+9^n) is converges

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sum from n=1 to infinity of 9/(6+9^n)

n = 1 \sum \infty \frac{9}{6 + 9 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{9}{6 + 9 ^{n}}

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty \frac{1}{6 + 9 ^{n}}

Apply Series Ratio Test:

converges

= 9 converges

= converges

x \to - 3 lim (\frac{x ^{2} - 5}{3 - x})

\int \frac{1}{1 - 9 x ^{2}} d x

\int 2 cos (x^{2}) d x

\frac{d}{d x} (\frac{2 x ^{2} + 8 x + 8}{x})

f (x) = (2 e^{x} - 3 x^{5})

What is the sum from n=1 to infinity of 9/(6+9^n) ?
The sum from n=1 to infinity of 9/(6+9^n) is converges