What is the sum from n=1 to infinity of-4(5/8)^n ?

The sum from n=1 to infinity of-4(5/8)^n is -20/3

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sum from n=1 to infinity of-4(5/8)^n

n = 1 \sum \infty - 4 (\frac{5}{8})^{n}

- \frac{20}{3}

Solution steps

n = 1 \sum \infty - 4 (\frac{5}{8})^{n}

Apply the constant multiplication rule:

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

= - 4 \cdot n = 1 \sum \infty (\frac{5}{8})^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= - 4 (n = 0 \sum \infty (\frac{5}{8})^{n} - (\frac{5}{8})^{0})

n = 0 \sum \infty (\frac{5}{8})^{n} = \frac{8}{3}

= - 4 (\frac{8}{3} - (\frac{5}{8})^{0})

Simplify

- 4 (\frac{8}{3} - (\frac{5}{8})^{0}) : - \frac{20}{3}

= - \frac{20}{3}

k = 1 \sum \infty \frac{9}{k} - \frac{9}{k + 1}

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

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

n = 0 \sum \infty \frac{1}{1 - 6 n}

n = 0 \sum \infty (0.2)^{n}

What is the sum from n=1 to infinity of-4(5/8)^n ?
The sum from n=1 to infinity of-4(5/8)^n is -20/3