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

The sum from n=1 to infinity of 2/(7^n) is 1/3

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sum from n=1 to infinity of 2/(7^n)

n = 1 \sum \infty \frac{2}{7 ^{n}}

\frac{1}{3}

Solution steps

n = 1 \sum \infty \frac{2}{7 ^{n}}

Apply the constant multiplication rule:

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

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

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

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

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

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

Simplify

2 (\frac{7}{6} - \frac{1}{7 ^{0}}) : \frac{1}{3}

= \frac{1}{3}

n = 0 \sum \infty (- 1 + \frac{4}{n})^{4}

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

n = 1 \sum \infty (\frac{2}{9})^{n}

n = 1 \sum \infty n e^{- 16 n}

n = 1 \sum \infty (- 0.005)^{n}

What is the sum from n=1 to infinity of 2/(7^n) ?
The sum from n=1 to infinity of 2/(7^n) is 1/3