What is the sum from n=0 to infinity of 10(-6/10)^n ?

The sum from n=0 to infinity of 10(-6/10)^n is 25/4

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sum from n=0 to infinity of 10(-6/10)^n

n = 0 \sum \infty 10 (- \frac{6}{10})^{n}

\frac{25}{4}

Solution steps

n = 0 \sum \infty 10 (- \frac{6}{10})^{n}

Apply the constant multiplication rule:

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

= 10 \cdot n = 0 \sum \infty (- \frac{6}{10})^{n}

Apply Series Geometric Test:

\frac{5}{8}

= 10 \cdot \frac{5}{8}

Simplify

10 \cdot \frac{5}{8} : \frac{25}{4}

= \frac{25}{4}

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

n = 2 \sum \infty n e^{- n^{2}}

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

n = 2 \sum \infty \frac{7}{5 ^{n}}

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

What is the sum from n=0 to infinity of 10(-6/10)^n ?
The sum from n=0 to infinity of 10(-6/10)^n is 25/4