What is the sum from n=2 to infinity of (1+1/4)^{-n} ?

The sum from n=2 to infinity of (1+1/4)^{-n} is converges

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sum from n=2 to infinity of (1+1/4)^{-n}

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

converges

Solution steps

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

Join

1 + \frac{1}{4} : \frac{5}{4}

= n = 2 \sum \infty (\frac{5}{4})^{- n}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty (- \frac{7}{8})^{n - 1}

n = 1 \sum \infty \frac{n + 1}{2 n - 1}

n = 0 \sum \infty (- 1)^{n - 1}

n = 1 \sum \infty (9.9)^{n}

n = 1 \sum \infty 7 (0.8)^{n - 1}

What is the sum from n=2 to infinity of (1+1/4)^{-n} ?
The sum from n=2 to infinity of (1+1/4)^{-n} is converges