What is the sum from n=1 to infinity of-1/(6^n) ?

The sum from n=1 to infinity of-1/(6^n) is -1/5

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sum from n=1 to infinity of-1/(6^n)

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

- \frac{1}{5}

Decimal

- 0.2

Solution steps

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

Apply the constant multiplication rule:

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

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

n = 0 \sum \infty \frac{1}{6 ^{n}} = \frac{6}{5}

= - (\frac{6}{5} - \frac{1}{6 ^{0}})

Simplify

- (\frac{6}{5} - \frac{1}{6 ^{0}}) : - \frac{1}{5}

= - \frac{1}{5}

What is the sum from n=1 to infinity of-1/(6^n) ?
The sum from n=1 to infinity of-1/(6^n) is -1/5