What is the sum from n=3 to infinity of (1/3)^n ?

The sum from n=3 to infinity of (1/3)^n is converges

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sum from n=3 to infinity of (1/3)^n

n = 3 \sum \infty (\frac{1}{3})^{n}

converges

Solution steps

n = 3 \sum \infty (\frac{1}{3})^{n}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty n! (- e)^{- 2 n}

n = 0 \sum \infty e^{\frac{- n ^{6}}{8}}

k = 2 \sum \infty 3 e^{- 2 k}

n = 1 \sum \infty 3 (\frac{2}{3})^{n}

n = 1 \sum \infty 60 (0.1)^{n}

What is the sum from n=3 to infinity of (1/3)^n ?
The sum from n=3 to infinity of (1/3)^n is converges