What is the sum from n=1 to infinity of (4^n)/(7^n) ?

The sum from n=1 to infinity of (4^n)/(7^n) is 4/3

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sum from n=1 to infinity of (4^n)/(7^n)

n = 1 \sum \infty \frac{4 ^{n}}{7 ^{n}}

\frac{4}{3}

Solution steps

n = 1 \sum \infty \frac{4 ^{n}}{7 ^{n}}

Apply exponent rule:

\frac{a ^{c}}{b ^{c}} = (\frac{a}{b})^{c}

= n = 1 \sum \infty (\frac{4}{7})^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty (\frac{4}{7})^{n} - (\frac{4}{7})^{0}

n = 0 \sum \infty (\frac{4}{7})^{n} = \frac{7}{3}

= \frac{7}{3} - (\frac{4}{7})^{0}

Simplify

\frac{7}{3} - (\frac{4}{7})^{0} : \frac{4}{3}

= \frac{4}{3}

\frac{d}{d x} (0.006 x^{3} + 0.07 x^{2} + 18 x + 600)

\int_{- \infty}^{\infty} \frac{2 x}{( x ^{2} + 1 ) ^{12}} d x

\frac{d y}{d x} = y (2 x - 5)

x \to 4 lim (\frac{x ^{2} - 8 x + 16}{x - 4})

cos (5 x)

What is the sum from n=1 to infinity of (4^n)/(7^n) ?
The sum from n=1 to infinity of (4^n)/(7^n) is 4/3