What is the sum from n=0 to infinity of 15/4 (1/4)^n ?

The sum from n=0 to infinity of 15/4 (1/4)^n is 5

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sum from n=0 to infinity of 15/4 (1/4)^n

n = 0 \sum \infty \frac{15}{4} (\frac{1}{4})^{n}

5

Solution steps

n = 0 \sum \infty \frac{15}{4} (\frac{1}{4})^{n}

Apply the constant multiplication rule:

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

= \frac{15}{4} \cdot n = 0 \sum \infty (\frac{1}{4})^{n}

Apply Series Geometric Test:

\frac{4}{3}

= \frac{15}{4} \cdot \frac{4}{3}

Simplify

\frac{15}{4} \cdot \frac{4}{3} : 5

= 5

n = 1 \sum \infty \frac{1}{2 ^{n} - 9}

n = 7 \sum \infty \frac{1}{n ( n + 1 )}

n = 1 \sum \infty \frac{7}{n ^{2} + 16}

n = 0 \sum \infty \frac{n}{7 n ^{3} + 1}

n = 0 \sum \infty ln (2 - n)

What is the sum from n=0 to infinity of 15/4 (1/4)^n ?
The sum from n=0 to infinity of 15/4 (1/4)^n is 5