What is the sum from n=1 to infinity of 5*0.02^{n-1} ?

The sum from n=1 to infinity of 5*0.02^{n-1} is converges

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sum from n=1 to infinity of 5*0.02^{n-1}

n = 1 \sum \infty 5 \cdot 0.0 2^{n - 1}

converges

Solution steps

n = 1 \sum \infty 5 \cdot 0.0 2^{n - 1}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty 0.0 2^{n - 1}

Apply Series Ratio Test:

converges

= 5 converges

= converges

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

n = 1 \sum \infty \frac{1}{7 n - 1}

k = 1 \sum \infty k^{3} e^{- 2 k}

n = 1 \sum \infty (- 1.99 + 1)^{n}

n = 1 \sum \infty \frac{2}{n ^{10}}

What is the sum from n=1 to infinity of 5*0.02^{n-1} ?
The sum from n=1 to infinity of 5*0.02^{n-1} is converges