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

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

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

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

4

Solution steps

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

\int_{7}^{8} \frac{7}{50 + x ^{2} - 14 x} d x

\int 2 x^{\frac{5}{2}} d x

x \to 5 lim (f (x))

y = x^{7 x}

e^{- 2 x}, 2

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