What is the sum from n=1 to infinity of (3^n)/(5^n) ?

The sum from n=1 to infinity of (3^n)/(5^n) is 3/2

sum from n=1 to infinity of (3^n)/(5^n)

n = 1 \sum \infty \frac{3 ^{n}}{5 ^{n}}

\frac{3}{2}

Decimal

1.5

Solution steps

n = 1 \sum \infty \frac{3 ^{n}}{5 ^{n}}

Apply exponent rule:

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

= n = 1 \sum \infty (\frac{3}{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{3}{5})^{n} - (\frac{3}{5})^{0}

n = 0 \sum \infty (\frac{3}{5})^{n} = \frac{5}{2}

= \frac{5}{2} - (\frac{3}{5})^{0}

Simplify

\frac{5}{2} - (\frac{3}{5})^{0} : \frac{3}{2}

= \frac{3}{2}

x \to 0 lim (\frac{9 + x}{x ^{3}})

\frac{\partial}{\partial x} ((2 x^{2} - 5 x)^{3} (2 x + 1))

\int_{0}^{27} 3 x d x

s \to \infty lim (c t)

a re a x = y, x = 2, y = 0

What is the sum from n=1 to infinity of (3^n)/(5^n) ?
The sum from n=1 to infinity of (3^n)/(5^n) is 3/2