What is the sum from n=1 to infinity of (7^n)/(10^n) ?

The sum from n=1 to infinity of (7^n)/(10^n) is 7/3

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sum from n=1 to infinity of (7^n)/(10^n)

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

\frac{7}{3}

Solution steps

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

Apply exponent rule:

\frac{a ^{c}}{b ^{c}} = (\frac{a}{b})^{c}

= n = 1 \sum \infty (\frac{7}{10})^{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{7}{10})^{n} - (\frac{7}{10})^{0}

n = 0 \sum \infty (\frac{7}{10})^{n} = \frac{10}{3}

= \frac{10}{3} - (\frac{7}{10})^{0}

Simplify

\frac{10}{3} - (\frac{7}{10})^{0} : \frac{7}{3}

= \frac{7}{3}

k = 5 \sum \infty 2 (- \frac{1}{4})^{k}

n = 4 \sum \infty 4 (6 - 2 n)

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

n = 0 \sum \infty \frac{9}{1 0 ^{n}}

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

What is the sum from n=1 to infinity of (7^n)/(10^n) ?
The sum from n=1 to infinity of (7^n)/(10^n) is 7/3