What is the sum from n=1 to infinity of (4^n)/(5^n) ?

The sum from n=1 to infinity of (4^n)/(5^n) is 4

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sum from n=1 to infinity of (4^n)/(5^n)

n = 1 \sum \infty \frac{4 ^{n}}{5 ^{n}}

4

Solution steps

n = 1 \sum \infty \frac{4 ^{n}}{5 ^{n}}

Apply exponent rule:

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

= n = 1 \sum \infty (\frac{4}{5})^{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}{5})^{n} - (\frac{4}{5})^{0}

n = 0 \sum \infty (\frac{4}{5})^{n} = 5

= 5 - (\frac{4}{5})^{0}

Simplify

= 4

8 t^{2} - t

\frac{d}{d x} (5 e^{x^{3}})

(3 x + 5)^{2}

(2 x - 1) d x + (7 y + 8) d y = 0

x \to - 5 lim (4)

What is the sum from n=1 to infinity of (4^n)/(5^n) ?
The sum from n=1 to infinity of (4^n)/(5^n) is 4