What is the sum from n=1 to infinity of (9.9)^n ?

The sum from n=1 to infinity of (9.9)^n is diverges

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sum from n=1 to infinity of (9.9)^n

n = 1 \sum \infty (9.9)^{n}

diverges

Solution steps

n = 1 \sum \infty 9. 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 9. 9^{n} - 9. 9^{0}

n = 0 \sum \infty 9. 9^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

n = 1 \sum \infty 7 (0.8)^{n - 1}

n = 1 \sum \infty \frac{n}{13 n + 4}

n = 0 \sum \infty \frac{4}{5 - n}

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

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

What is the sum from n=1 to infinity of (9.9)^n ?
The sum from n=1 to infinity of (9.9)^n is diverges