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

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

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

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

diverges

Solution steps

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

Apply exponent rule:

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

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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x \to π lim ((2 x - 2 π) cot (x))

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\frac{\partial}{\partial x} (\frac{6 x}{x ^{2} + y ^{2}})

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