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

The sum from n=2 to infinity of [ 2/(5^n)] is 1/10

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

n = 2 \sum \infty [\frac{2}{5 ^{n}}]

\frac{1}{10}

Solution steps

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

Apply the constant multiplication rule:

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

= 2 \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}

= 2 (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}

= 2 (\frac{5}{4} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}})

Simplify

2 (\frac{5}{4} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}}) : \frac{1}{10}

= \frac{1}{10}

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

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

n = 0 \sum \infty (\frac{4}{3})^{1 - n}

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

n = 1 \sum \infty 10 (- \frac{7}{8})^{n}

What is the sum from n=2 to infinity of [ 2/(5^n)] ?
The sum from n=2 to infinity of [ 2/(5^n)] is 1/10