What is the sum from n=1 to infinity of (3^n)/(8^n) ?

The sum from n=1 to infinity of (3^n)/(8^n) is 3/5

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sum from n=1 to infinity of (3^n)/(8^n)

n = 1 \sum \infty \frac{3 ^{n}}{8 ^{n}}

\frac{3}{5}

Solution steps

n = 1 \sum \infty \frac{3 ^{n}}{8 ^{n}}

Apply exponent rule:

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

= n = 1 \sum \infty (\frac{3}{8})^{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{3}{8})^{n} - (\frac{3}{8})^{0}

n = 0 \sum \infty (\frac{3}{8})^{n} = \frac{8}{5}

= \frac{8}{5} - (\frac{3}{8})^{0}

Simplify

\frac{8}{5} - (\frac{3}{8})^{0} : \frac{3}{5}

= \frac{3}{5}

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

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

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

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

n = 0 \sum \infty \frac{i ^{2 n}}{2 n}

What is the sum from n=1 to infinity of (3^n)/(8^n) ?
The sum from n=1 to infinity of (3^n)/(8^n) is 3/5