What is the sum from n=5 to infinity of 1/(27^n) ?

The sum from n=5 to infinity of 1/(27^n) is converges

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sum from n=5 to infinity of 1/(27^n)

n = 5 \sum \infty \frac{1}{2 7 ^{n}}

converges

Solution steps

n = 5 \sum \infty \frac{1}{2 7 ^{n}}

Apply Series Ratio Test:

converges

= converges

x \to 0 lim (sin (sin (x)))

\int \frac{sin ( x )}{1 + cos ^{2} ( x )} d x

\int_{7}^{9} (x^{2} + 2 x - 3) d x

y^{^{''}} + 6 y^{^{'}} + 9 y = 32 e^{t}, y (0) = 1, y^{^{'}} (0) = 7

\frac{d}{d x} ((x^{2} + 1)^{x})

What is the sum from n=5 to infinity of 1/(27^n) ?
The sum from n=5 to infinity of 1/(27^n) is converges