What is the sum from m=2 to infinity of 2/(3^m) ?

The sum from m=2 to infinity of 2/(3^m) is 1/3

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sum from m=2 to infinity of 2/(3^m)

m = 2 \sum \infty \frac{2}{3 ^{m}}

\frac{1}{3}

Solution steps

m = 2 \sum \infty \frac{2}{3 ^{m}}

Apply the constant multiplication rule:

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

= 2 \cdot m = 2 \sum \infty \frac{1}{3 ^{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}{3 ^{m}} - \frac{1}{3 ^{0}} - \frac{1}{3 ^{1}})

m = 0 \sum \infty \frac{1}{3 ^{m}} = \frac{3}{2}

= 2 (\frac{3}{2} - \frac{1}{3 ^{0}} - \frac{1}{3 ^{1}})

Simplify

2 (\frac{3}{2} - \frac{1}{3 ^{0}} - \frac{1}{3 ^{1}}) : \frac{1}{3}

= \frac{1}{3}

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

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

n = 1 \sum \infty (\frac{e}{π})^{n + 1}

n = 1 \sum \infty \frac{1}{( n + 3 ) ^{3}}

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

What is the sum from m=2 to infinity of 2/(3^m) ?
The sum from m=2 to infinity of 2/(3^m) is 1/3