What is the sum from n=2 to infinity of 4/(5^n) ?

The sum from n=2 to infinity of 4/(5^n) is 1/5

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sum from n=2 to infinity of 4/(5^n)

n = 2 \sum \infty \frac{4}{5 ^{n}}

\frac{1}{5}

Decimal

0.2

Solution steps

n = 2 \sum \infty \frac{4}{5 ^{n}}

Apply the constant multiplication rule:

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

= 4 \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}

= 4 (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}

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

Simplify

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

= \frac{1}{5}

What is the sum from n=2 to infinity of 4/(5^n) ?
The sum from n=2 to infinity of 4/(5^n) is 1/5