What is the sum from n=1 to infinity of 1/24 (3/4)^n ?

The sum from n=1 to infinity of 1/24 (3/4)^n is 1/8

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sum from n=1 to infinity of 1/24 (3/4)^n

n = 1 \sum \infty \frac{1}{24} (\frac{3}{4})^{n}

\frac{1}{8}

Solution steps

n = 1 \sum \infty \frac{1}{24} (\frac{3}{4})^{n}

Apply the constant multiplication rule:

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

= \frac{1}{24} \cdot n = 1 \sum \infty (\frac{3}{4})^{n}

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

= \frac{1}{24} (n = 0 \sum \infty (\frac{3}{4})^{n} - (\frac{3}{4})^{0})

n = 0 \sum \infty (\frac{3}{4})^{n} = 4

= \frac{1}{24} (4 - (\frac{3}{4})^{0})

Simplify

\frac{1}{24} (4 - (\frac{3}{4})^{0}) : \frac{1}{8}

= \frac{1}{8}

n = 0 \sum \infty A n \cdot (- 1)^{n}

k = 0 \sum \infty \frac{1}{k ^{2} + 4}

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

n = 0 \sum \infty (\frac{n}{4 - n})^{3}

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

What is the sum from n=1 to infinity of 1/24 (3/4)^n ?
The sum from n=1 to infinity of 1/24 (3/4)^n is 1/8