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

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

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

n = 0 \sum \infty \frac{1}{5} (\frac{4}{5})^{n}

1

Solution steps

n = 0 \sum \infty \frac{1}{5} (\frac{4}{5})^{n}

Apply the constant multiplication rule:

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

= \frac{1}{5} \cdot n = 0 \sum \infty (\frac{4}{5})^{n}

Apply Series Geometric Test:

5

= \frac{1}{5} \cdot 5

Simplify

\frac{1}{5} \cdot 5 : 1

= 1

n = 1 \sum \infty \frac{1}{n ^{2.5}}

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

n = 1 \sum \infty \frac{6}{4 n ^{2} - 1}

n = 0 \sum \infty 4 - \frac{1}{n ( n + 1 )}

k = 1 \sum \infty k e^{- 3 k}

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