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

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

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

n = 4 \sum \infty \frac{1}{2 ^{n}}

converges

Solution steps

n = 4 \sum \infty \frac{1}{2 ^{n}}

Apply Series Ratio Test:

converges

= converges

\int_{1}^{6} \frac{x ^{2} + 2}{7 x - x ^{2}} d x

\int \frac{1}{sec ( θ )} d θ

x + x (y^{^{'}}) 2 - 1 = 0

y^{^{''}} + 2 y^{^{'}} = 3

\frac{d}{d x} (y - 3 x)

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