What is the sum from n=1 to infinity of 4/(7^n) ?

The sum from n=1 to infinity of 4/(7^n) is 2/3

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sum from n=1 to infinity of 4/(7^n)

n = 1 \sum \infty \frac{4}{7 ^{n}}

\frac{2}{3}

Solution steps

n = 1 \sum \infty \frac{4}{7 ^{n}}

Apply the constant multiplication rule:

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

= 4 \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}

= 4 (n = 0 \sum \infty \frac{1}{7 ^{n}} - \frac{1}{7 ^{0}})

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

= 4 (\frac{7}{6} - \frac{1}{7 ^{0}})

Simplify

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

= \frac{2}{3}

n = 0 \sum \infty (- \frac{9}{10})^{n}

n = 1 \sum \infty \frac{3}{2 ^{n - 1}}

n = 1 \sum \infty \frac{3}{2 ^{2 n - 1}}

n = 0 \sum \infty 5 (- \frac{2}{3})^{n}

n = 1 \sum \infty \frac{1}{3 + e ^{- n}}

What is the sum from n=1 to infinity of 4/(7^n) ?
The sum from n=1 to infinity of 4/(7^n) is 2/3