What is the sum from n=8 to infinity of 1/(4^n) ?

The sum from n=8 to infinity of 1/(4^n) is converges

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sum from n=8 to infinity of 1/(4^n)

n = 8 \sum \infty \frac{1}{4 ^{n}}

converges

Solution steps

n = 8 \sum \infty \frac{1}{4 ^{n}}

Apply Series Ratio Test:

converges

= converges

\int \frac{x ^{3}}{( x + 1 ) ^{3}} d x

\frac{d}{d x} (2 x (x + 4)^{3})

\frac{\partial}{\partial x} (x^{2} + 3 y^{2} - 6 y + 4 x y)

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

y = - 3 x^{2}

What is the sum from n=8 to infinity of 1/(4^n) ?
The sum from n=8 to infinity of 1/(4^n) is converges