What is the sum from n=1 to infinity of 7(0.1)^{n-1} ?

The sum from n=1 to infinity of 7(0.1)^{n-1} is converges

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sum from n=1 to infinity of 7(0.1)^{n-1}

n = 1 \sum \infty 7 (0.1)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 7 \cdot 0. 1^{n - 1}

Apply the constant multiplication rule:

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

= 7 \cdot n = 1 \sum \infty 0. 1^{n - 1}

Apply Series Ratio Test:

converges

= 7 converges

= converges

n = 1 \sum \infty \frac{8}{6 n ^{2} + 6 n}

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

n = 2 \sum \infty \frac{7}{3 ^{n}}

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

n = 0 \sum \infty \frac{n}{6 + n ^{4}}

What is the sum from n=1 to infinity of 7(0.1)^{n-1} ?
The sum from n=1 to infinity of 7(0.1)^{n-1} is converges