What is the sum from n=0 to infinity of 0.52(0.01)^n ?

The sum from n=0 to infinity of 0.52(0.01)^n is 0.52525…

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sum from n=0 to infinity of 0.52(0.01)^n

n = 0 \sum \infty 0.52 (0.01)^{n}

0.52525 \dots

Solution steps

n = 0 \sum \infty 0.52 \cdot 0.0 1^{n}

Apply the constant multiplication rule:

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

= 0.52 \cdot n = 0 \sum \infty 0.0 1^{n}

Apply Series Geometric Test:

1.01010 \dots

= 0.52 \cdot 1.01010 \dots

Simplify

= 0.52525 \dots

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

n = 0 \sum \infty (\frac{13}{3})^{n}

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

n = 1 \sum \infty \frac{3 n !}{2 n + 1}

n = 0 \sum \infty ln (3 + n)

What is the sum from n=0 to infinity of 0.52(0.01)^n ?
The sum from n=0 to infinity of 0.52(0.01)^n is 0.52525…