What is the sum from n=1 to infinity of (5^n)/(7^n) ?

The sum from n=1 to infinity of (5^n)/(7^n) is 5/2

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sum from n=1 to infinity of (5^n)/(7^n)

n = 1 \sum \infty \frac{5 ^{n}}{7 ^{n}}

\frac{5}{2}

Solution steps

n = 1 \sum \infty \frac{5 ^{n}}{7 ^{n}}

Apply exponent rule:

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

= n = 1 \sum \infty (\frac{5}{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{5}{7})^{n} - (\frac{5}{7})^{0}

n = 0 \sum \infty (\frac{5}{7})^{n} = \frac{7}{2}

= \frac{7}{2} - (\frac{5}{7})^{0}

Simplify

\frac{7}{2} - (\frac{5}{7})^{0} : \frac{5}{2}

= \frac{5}{2}

n = 0 \sum \infty \frac{2}{( 2 n + 1 ) ^{2}}

x = 0 \sum \infty x

n = 0 \sum \infty 3 n - 3

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

n = 0 \sum \infty \frac{n + 1}{n ^{3}}

What is the sum from n=1 to infinity of (5^n)/(7^n) ?
The sum from n=1 to infinity of (5^n)/(7^n) is 5/2