What is the sum from n=1 to infinity of 9(0.9)^{n-1} ?

The sum from n=1 to infinity of 9(0.9)^{n-1} is converges

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sum from n=1 to infinity of 9(0.9)^{n-1}

n = 1 \sum \infty 9 (0.9)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 9 \cdot 0. 9^{n - 1}

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty 0. 9^{n - 1}

Apply Series Ratio Test:

converges

= 9 converges

= converges

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

n = 0 \sum \infty \frac{2}{n ^{3} + 5}

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

k = 6 \sum \infty \frac{1}{( k - 5 ) ^{4}}

n = 1 \sum \infty 53 \cdot (\frac{1}{2})^{n}

What is the sum from n=1 to infinity of 9(0.9)^{n-1} ?
The sum from n=1 to infinity of 9(0.9)^{n-1} is converges