What is the sum from n=1 to infinity of (4^n)/(6^n) ?

The sum from n=1 to infinity of (4^n)/(6^n) is 2

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sum from n=1 to infinity of (4^n)/(6^n)

n = 1 \sum \infty \frac{4 ^{n}}{6 ^{n}}

2

Solution steps

n = 1 \sum \infty \frac{4 ^{n}}{6 ^{n}}

Simplify

\frac{4 ^{n}}{6 ^{n}} : (\frac{2}{3})^{n}

= n = 1 \sum \infty (\frac{2}{3})^{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{2}{3})^{n} - (\frac{2}{3})^{0}

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

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

Simplify

= 2

\int (((x)^{2} - 5)^{5} 2 x) d x

x \to - 1 + lim (ln ∣ x + 1 ∣)

\frac{d}{d x} (x^{2} e^{\frac{- x}{10}})

f (x) = \frac{x}{x ^{2} - 15}, at x = 4

y^{^{''}} + 25 y^{^{'}} = 0

What is the sum from n=1 to infinity of (4^n)/(6^n) ?
The sum from n=1 to infinity of (4^n)/(6^n) is 2