What is the sum from n=0 to infinity of 1/4 (3/4)^n ?

The sum from n=0 to infinity of 1/4 (3/4)^n is 1

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sum from n=0 to infinity of 1/4 (3/4)^n

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

1

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

4

= \frac{1}{4} \cdot 4

Simplify

\frac{1}{4} \cdot 4 : 1

= 1

n = 1 \sum \infty 12 sin (\frac{1}{n})

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

n = 0 \sum \infty n^{- \frac{3}{2}}

n = 3 \sum \infty \frac{1}{( n - 2 ) ^{4}}

n = 1 \sum \infty \frac{9}{7 ^{n}}

What is the sum from n=0 to infinity of 1/4 (3/4)^n ?
The sum from n=0 to infinity of 1/4 (3/4)^n is 1