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

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

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

n = 5 \sum \infty \frac{1}{2 ^{n}}

converges

Solution steps

n = 5 \sum \infty \frac{1}{2 ^{n}}

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (7 x^{2} ln (x))

x^{3} + y^{3} - 3 x^{2} y^{2} + 1

y^{^{''}} + y^{^{'}} - 2 y = 2 t

\int_{0}^{3} x^{2} d x

\int \frac{2 x ^{3} + x ^{2} - x}{x ^{2}} d x

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