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

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

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

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

\frac{35}{4}

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

\frac{5}{4}

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

Simplify

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

= \frac{35}{4}

\frac{d}{d x} (\frac{x}{cos ( x )})

\frac{\partial}{\partial x} (x e^{4 x + 2 y})

\int \frac{1}{6 + 2 x} d x

\frac{d}{d x} ((x^{5} + 2)^{2} (x^{3} + 4)^{4})

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

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