What is the sum from n=0 to infinity of 9/(5^n) ?

The sum from n=0 to infinity of 9/(5^n) is 45/4

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sum from n=0 to infinity of 9/(5^n)

n = 0 \sum \infty \frac{9}{5 ^{n}}

\frac{45}{4}

Solution steps

n = 0 \sum \infty \frac{9}{5 ^{n}}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 9 \cdot n = 0 \sum \infty \frac{1}{5 ^{n}}

Apply Series Geometric Test:

\frac{5}{4}

= 9 \cdot \frac{5}{4}

Simplify

9 \cdot \frac{5}{4} : \frac{45}{4}

= \frac{45}{4}

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

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

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

k = 4 \sum \infty \frac{6}{k} - \frac{6}{k + 1}

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

What is the sum from n=0 to infinity of 9/(5^n) ?
The sum from n=0 to infinity of 9/(5^n) is 45/4