What is the sum from n=0 to infinity of 3^n7^{-n} ?

The sum from n=0 to infinity of 3^n7^{-n} is 7/4

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sum from n=0 to infinity of 3^n7^{-n}

n = 0 \sum \infty 3^{n} 7^{- n}

\frac{7}{4}

Solution steps

n = 0 \sum \infty 3^{n} \cdot 7^{- n}

Simplify

7^{- n} : \frac{1}{7 ^{n}}

= n = 0 \sum \infty 3^{n} \cdot \frac{1}{7 ^{n}}

Simplify

3^{n} \cdot \frac{1}{7 ^{n}} : (\frac{3}{7})^{n}

= n = 0 \sum \infty (\frac{3}{7})^{n}

Apply Series Geometric Test:

\frac{7}{4}

= \frac{7}{4}

\frac{\partial}{\partial y} (- x y^{2})

y = 2 x e^{- 6 x} - 4 e^{x^{3}}

f (x) = \frac{10 x}{x ^{2} - 6}, (4, 4)

4 x^{2} - 3

What is the sum from n=0 to infinity of 3^n7^{-n} ?
The sum from n=0 to infinity of 3^n7^{-n} is 7/4