What is the sum from n=0 to infinity of (4/11)^n ?

The sum from n=0 to infinity of (4/11)^n is 11/7

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sum from n=0 to infinity of (4/11)^n

n = 0 \sum \infty (\frac{4}{11})^{n}

\frac{11}{7}

Solution steps

n = 0 \sum \infty (\frac{4}{11})^{n}

Apply Series Geometric Test:

\frac{11}{7}

= \frac{11}{7}

j = 1 \sum \infty 2 (- \frac{3}{5})^{j}

n = 1 \sum \infty - \frac{1}{1 0 ^{n}}

n = 1 \sum \infty 4 (\frac{π}{4})^{n}

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

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

What is the sum from n=0 to infinity of (4/11)^n ?
The sum from n=0 to infinity of (4/11)^n is 11/7