What is the sum from n=2 to infinity of (1+1/3)^{-n} ?

The sum from n=2 to infinity of (1+1/3)^{-n} is converges

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sum from n=2 to infinity of (1+1/3)^{-n}

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

converges

Solution steps

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

Join

1 + \frac{1}{3} : \frac{4}{3}

= n = 2 \sum \infty (\frac{4}{3})^{- n}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty sin (\frac{nπ}{2})

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

n = 0 \sum \infty \frac{1}{100 + n}

n = 1 \sum \infty \frac{7}{n ( n + 2 )}

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

What is the sum from n=2 to infinity of (1+1/3)^{-n} ?
The sum from n=2 to infinity of (1+1/3)^{-n} is converges