What is the sum from n=2 to infinity of | 3/(5^n)| ?

The sum from n=2 to infinity of | 3/(5^n)| is 3/20

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sum from n=2 to infinity of | 3/(5^n)|

n = 2 \sum \infty \frac{3}{5 ^{n}}

\frac{3}{20}

Solution steps

n = 2 \sum \infty \frac{3}{5 ^{n}}

Simplify

\frac{3}{5 ^{n}} : \frac{3}{5 ^{n}}

= n = 2 \sum \infty \frac{3}{5 ^{n}}

Apply the constant multiplication rule:

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

= 3 \cdot n = 2 \sum \infty \frac{1}{5 ^{n}}

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

= 3 (n = 0 \sum \infty \frac{1}{5 ^{n}} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}})

n = 0 \sum \infty \frac{1}{5 ^{n}} = \frac{5}{4}

= 3 (\frac{5}{4} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}})

Simplify

3 (\frac{5}{4} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}}) : \frac{3}{20}

= \frac{3}{20}

n = 0 \sum \infty 5 (\frac{2}{2})^{9}

n = 1 \sum \infty 4^{n} 5^{- n}

x = 1 \sum \infty \frac{2}{x ^{2} + 4 x + 3}

n = 0 \sum \infty \frac{1}{cos ( \frac{1}{n} )}

n = 1 \sum \infty \frac{1}{4 n ^{2} - 81}

What is the sum from n=2 to infinity of | 3/(5^n)| ?
The sum from n=2 to infinity of | 3/(5^n)| is 3/20