What is the sum from k=0 to infinity of (1/7)^k ?

The sum from k=0 to infinity of (1/7)^k is 7/6

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sum from k=0 to infinity of (1/7)^k

k = 0 \sum \infty (\frac{1}{7})^{k}

\frac{7}{6}

Solution steps

k = 0 \sum \infty (\frac{1}{7})^{k}

Apply Series Geometric Test:

\frac{7}{6}

= \frac{7}{6}

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

n = 0 \sum \infty (- 1)^{n} 5^{n}

n = 1 \sum \infty \frac{1}{6 n + 1}

n = 0 \sum \infty (\frac{8 π}{e})^{n}

n = 1 \sum \infty (cos (20))^{n}

What is the sum from k=0 to infinity of (1/7)^k ?
The sum from k=0 to infinity of (1/7)^k is 7/6