What is the sum from n=1 to infinity of-1/(13^n) ?

The sum from n=1 to infinity of-1/(13^n) is -1/12

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sum from n=1 to infinity of-1/(13^n)

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

- \frac{1}{12}

Solution steps

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

Apply the constant multiplication rule:

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

= - n = 1 \sum \infty \frac{1}{1 3 ^{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 3 ^{n}} - \frac{1}{1 3 ^{0}})

n = 0 \sum \infty \frac{1}{1 3 ^{n}} = \frac{13}{12}

= - (\frac{13}{12} - \frac{1}{1 3 ^{0}})

Simplify

- (\frac{13}{12} - \frac{1}{1 3 ^{0}}) : - \frac{1}{12}

= - \frac{1}{12}

n = 0 \sum \infty \frac{1}{n} + \frac{1}{n + 1}

j = 1 \sum \infty 5^{- 3 j}

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

n = 1 \sum \infty sin^{n} (1)

n = 1 \sum \infty \frac{n ^{6}}{n !}

What is the sum from n=1 to infinity of-1/(13^n) ?
The sum from n=1 to infinity of-1/(13^n) is -1/12