What is the sum from n=1 to infinity of (9^n)/(10^n) ?

The sum from n=1 to infinity of (9^n)/(10^n) is 9

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sum from n=1 to infinity of (9^n)/(10^n)

n = 1 \sum \infty \frac{9 ^{n}}{1 0 ^{n}}

9

Solution steps

n = 1 \sum \infty \frac{9 ^{n}}{1 0 ^{n}}

Apply exponent rule:

\frac{a ^{c}}{b ^{c}} = (\frac{a}{b})^{c}

= n = 1 \sum \infty (\frac{9}{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{9}{10})^{n} - (\frac{9}{10})^{0}

n = 0 \sum \infty (\frac{9}{10})^{n} = 10

= 10 - (\frac{9}{10})^{0}

Simplify

= 9

\int sec^{3} (8 x) d x

\int x^{2} 2 x + 5 d x

y = \frac{3}{x}, (4, \frac{3}{2})

t^{2} cos (t)

y^{^{''}} + 4 y^{^{'}} + 4 y = t + e^{2 t}

What is the sum from n=1 to infinity of (9^n)/(10^n) ?
The sum from n=1 to infinity of (9^n)/(10^n) is 9