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

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

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

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

\frac{1}{6}

Solution steps

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

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

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

n = 0 \sum \infty (\frac{1}{7})^{n} = \frac{7}{6}

= \frac{7}{6} - (\frac{1}{7})^{0}

Simplify

\frac{7}{6} - (\frac{1}{7})^{0} : \frac{1}{6}

= \frac{1}{6}

4 x - x^{2}, x^{2}, 2, 0

x \to - 6 - lim (\frac{1}{( x + 6 ) ^{6}})

f (x) = e^{x} + x

\frac{d y}{d x} = \frac{1 + cos ( 4 x )}{y e ^{y^{2}}}

\int (- \frac{1}{x ^{2}}) d x

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