What is the sum from n=1 to infinity of 5/(7^n) ?

The sum from n=1 to infinity of 5/(7^n) is 5/6

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sum from n=1 to infinity of 5/(7^n)

n = 1 \sum \infty \frac{5}{7 ^{n}}

\frac{5}{6}

Solution steps

n = 1 \sum \infty \frac{5}{7 ^{n}}

Apply the constant multiplication rule:

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

= 5 \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}

= 5 (n = 0 \sum \infty \frac{1}{7 ^{n}} - \frac{1}{7 ^{0}})

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

= 5 (\frac{7}{6} - \frac{1}{7 ^{0}})

Simplify

5 (\frac{7}{6} - \frac{1}{7 ^{0}}) : \frac{5}{6}

= \frac{5}{6}

y (x + y + 1) d x + 2 x y d y = 0

\int_{0}^{4} \frac{x}{x ^{2} + 1} d x

\frac{d}{d x} (x^{\frac{4}{5}} - x^{\frac{2}{3}})

\int \frac{2 x}{x + 3} d x

\frac{d}{d x} (ln (\frac{5 x ^{3} sin ( - 5 x )}{3 - x}))

What is the sum from n=1 to infinity of 5/(7^n) ?
The sum from n=1 to infinity of 5/(7^n) is 5/6