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))

f (x) = - arctan (x)

s im pl i f y x 5^{2 x}

\frac{d}{d x} \frac{1}{1 - sin ^{2} ( x )}

d er i v a t i v e (2 t - 1) (5 t - 5)^{- 1}

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