What is the sum from n=1 to infinity of (ln(3))^n ?

The sum from n=1 to infinity of (ln(3))^n is diverges

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sum from n=1 to infinity of (ln(3))^n

n = 1 \sum \infty (ln (3))^{n}

diverges

Solution steps

n = 1 \sum \infty (ln (3))^{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 (ln (3))^{n} - (ln (3))^{0}

n = 0 \sum \infty (ln (3))^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

f (x) = 7 x + 11, (2, 5)

y^{^{''}} - 4 y = e^{x} cos (x), y (0) = 0.7, y^{^{'}} (0) = - 1.5

\frac{d y}{d t} = y^{2}

f (x) = ln (6 x)

\int e^{2 x} cos (5 x) d x

What is the sum from n=1 to infinity of (ln(3))^n ?
The sum from n=1 to infinity of (ln(3))^n is diverges