What is the sum from n=0 to infinity of 7/(5^n) ?

The sum from n=0 to infinity of 7/(5^n) is 35/4

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sum from n=0 to infinity of 7/(5^n)

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

\frac{35}{4}

Decimal

8.75

Solution steps

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

Apply the constant multiplication rule:

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

= 7 \cdot n = 0 \sum \infty \frac{1}{5 ^{n}}

Apply Series Geometric Test:

\frac{5}{4}

= 7 \cdot \frac{5}{4}

Simplify

7 \cdot \frac{5}{4} : \frac{35}{4}

= \frac{35}{4}

What is the sum from n=0 to infinity of 7/(5^n) ?
The sum from n=0 to infinity of 7/(5^n) is 35/4