What is the sum from n=2 to infinity of 7(-3/8)^n ?

The sum from n=2 to infinity of 7(-3/8)^n is converges

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sum from n=2 to infinity of 7(-3/8)^n

n = 2 \sum \infty 7 (- \frac{3}{8})^{n}

converges

Solution steps

n = 2 \sum \infty 7 (- \frac{3}{8})^{n}

Apply the constant multiplication rule:

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

= 7 \cdot n = 2 \sum \infty (- \frac{3}{8})^{n}

Apply Series Ratio Test:

converges

= 7 converges

= converges

n = 2 \sum \infty (\frac{3}{5})^{4 n}

n = 0 \sum \infty 8 (\frac{1}{10})^{n}

n = 1 \sum \infty 4 n

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

n = 0 \sum \infty \frac{n}{1 + 8 n ^{3}}

What is the sum from n=2 to infinity of 7(-3/8)^n ?
The sum from n=2 to infinity of 7(-3/8)^n is converges