What is the sum from m=2 to infinity of 5/(6^m) ?

The sum from m=2 to infinity of 5/(6^m) is 1/6

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sum from m=2 to infinity of 5/(6^m)

m = 2 \sum \infty \frac{5}{6 ^{m}}

\frac{1}{6}

Solution steps

m = 2 \sum \infty \frac{5}{6 ^{m}}

Apply the constant multiplication rule:

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

= 5 \cdot m = 2 \sum \infty \frac{1}{6 ^{m}}

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

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

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

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

Simplify

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

= \frac{1}{6}

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What is the sum from m=2 to infinity of 5/(6^m) ?
The sum from m=2 to infinity of 5/(6^m) is 1/6