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

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

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

\frac{3}{2}

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}

\int x (x - 2)^{3} d x

f (x) = 3 + 4 x^{2} - 2 x^{3}, (2, 3)

\int (1 - x^{2})^{3} d x

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

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

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