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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 3 converges

= converges

\int 1 - 2 x d x

\int sin (x) + sinh (x) d x

\frac{d y}{d x} = \frac{( 10 + 15 x )}{x y ^{2}}, y (1) = 3

\frac{\partial}{\partial x} (x \cdot y)

\int \frac{3 x ^{2}}{1 + x ^{3}} d x

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