What is the sum from m=2 to infinity of 2/(7^m) ?

The sum from m=2 to infinity of 2/(7^m) is 1/21

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sum from m=2 to infinity of 2/(7^m)

m = 2 \sum \infty \frac{2}{7 ^{m}}

\frac{1}{21}

Solution steps

m = 2 \sum \infty \frac{2}{7 ^{m}}

Apply the constant multiplication rule:

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

= 2 \cdot m = 2 \sum \infty \frac{1}{7 ^{m}}

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

= 2 (m = 0 \sum \infty \frac{1}{7 ^{m}} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}})

m = 0 \sum \infty \frac{1}{7 ^{m}} = \frac{7}{6}

= 2 (\frac{7}{6} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}})

Simplify

2 (\frac{7}{6} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}}) : \frac{1}{21}

= \frac{1}{21}

n = 0 \sum \infty \frac{- 2}{n + 1}

n = 1 \sum \infty (- \frac{7}{9})^{n - 1}

n = 1 \sum \infty \frac{1}{5 + n ^{2}}

n = 2 \sum \infty \frac{3}{n ln ( n )}

n = 1 \sum \infty \frac{2}{( 4 n ^{2} - 1 )}

What is the sum from m=2 to infinity of 2/(7^m) ?
The sum from m=2 to infinity of 2/(7^m) is 1/21