What is the sum from n=0 to infinity of (3/4)^{-n} ?

The sum from n=0 to infinity of (3/4)^{-n} is diverges

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sum from n=0 to infinity of (3/4)^{-n}

n = 0 \sum \infty (\frac{3}{4})^{- n}

diverges

Solution steps

n = 0 \sum \infty (\frac{3}{4})^{- n}

Apply Series Ratio Test:

diverges

= diverges

n = 0 \sum \infty \frac{1}{ln ( 1 + n )}

n = 1 \sum \infty \frac{1}{4 + 3 ^{n}}

k = 1 \sum \infty \frac{1}{2 ^{k + 1}}

n = 0 \sum \infty \frac{4}{9 ^{n}}

n = 0 \sum \infty \frac{n}{n ^{2} + n + 1}

What is the sum from n=0 to infinity of (3/4)^{-n} ?
The sum from n=0 to infinity of (3/4)^{-n} is diverges