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

The sum from n=1 to infinity of (2/10)^n is 1/4

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sum from n=1 to infinity of (2/10)^n

n = 1 \sum \infty (\frac{2}{10})^{n}

\frac{1}{4}

Solution steps

n = 1 \sum \infty (\frac{2}{10})^{n}

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

= n = 0 \sum \infty (\frac{2}{10})^{n} - (\frac{2}{10})^{0}

n = 0 \sum \infty (\frac{2}{10})^{n} = \frac{5}{4}

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

Simplify

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

= \frac{1}{4}

n = 0 \sum \infty \frac{2}{3 n + 5}

n = 1 \sum \infty \frac{6 n}{e ^{n}}

n = 1 \sum \infty n

n = 0 \sum \infty \frac{1}{n 3 ^{n}}

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

What is the sum from n=1 to infinity of (2/10)^n ?
The sum from n=1 to infinity of (2/10)^n is 1/4