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

The sum from n=1 to infinity of (e/(10))^n is (10)/(10-e)-1

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

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

\frac{10}{10 - e} - 1

Solution steps

n = 1 \sum \infty (\frac{e}{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{e}{10})^{n} - (\frac{e}{10})^{0}

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

= \frac{10}{10 - e} - (\frac{e}{10})^{0}

Simplify

= \frac{10}{10 - e} - 1

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