What is the sum from n=1 to infinity of (ln(1.6))^n ?

The sum from n=1 to infinity of (ln(1.6))^n is 0.88680…

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sum from n=1 to infinity of (ln(1.6))^n

n = 1 \sum \infty (ln (1.6))^{n}

0.88680 \dots

Solution steps

n = 1 \sum \infty (ln (1.6))^{n}

Simplify

ln (1.6) : 0.47000 \dots

= n = 1 \sum \infty 0.47000 \dots^{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 0.47000 \dots^{n} - 0.47000 \dots^{0}

n = 0 \sum \infty 0.47000 \dots^{n} = 1.88680 \dots

= 1.88680 \dots - 0.47000 \dots^{0}

Simplify

= 0.88680 \dots

n = 0 \sum \infty \frac{n}{7 + n}

n = 0 \sum \infty (\frac{1}{18})^{n}

n = 0 \sum \infty n sin (\frac{π}{n})

k = 2 \sum \infty 5 (\frac{1}{3})^{k}

g = 1 \sum \infty g^{gx}

What is the sum from n=1 to infinity of (ln(1.6))^n ?
The sum from n=1 to infinity of (ln(1.6))^n is 0.88680…