What is the sum from n=1 to infinity of-1/(18^n) ?

The sum from n=1 to infinity of-1/(18^n) is -1/17

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sum from n=1 to infinity of-1/(18^n)

n = 1 \sum \infty - \frac{1}{1 8 ^{n}}

- \frac{1}{17}

Solution steps

n = 1 \sum \infty - \frac{1}{1 8 ^{n}}

Apply the constant multiplication rule:

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

= - n = 1 \sum \infty \frac{1}{1 8 ^{n}}

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

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

n = 0 \sum \infty \frac{1}{1 8 ^{n}} = \frac{18}{17}

= - (\frac{18}{17} - \frac{1}{1 8 ^{0}})

Simplify

- (\frac{18}{17} - \frac{1}{1 8 ^{0}}) : - \frac{1}{17}

= - \frac{1}{17}

n = 1 \sum \infty (cos (14))^{n}

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

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

n = 0 \sum \infty (\frac{5}{2})^{n - 1}

k = 0 \sum \infty (0.4)^{k}

What is the sum from n=1 to infinity of-1/(18^n) ?
The sum from n=1 to infinity of-1/(18^n) is -1/17