What is the sum from n=1 to infinity of (ln(2))/(n!) ?

The sum from n=1 to infinity of (ln(2))/(n!) is converges

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sum from n=1 to infinity of (ln(2))/(n!)

n = 1 \sum \infty \frac{ln ( 2 )}{n !}

converges

Solution steps

n = 1 \sum \infty \frac{ln ( 2 )}{n !}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= ln (2) \cdot n = 1 \sum \infty \frac{1}{n !}

Apply Series Ratio Test:

converges

= ln (2) converges

= converges

n = 0 \sum \infty \frac{n + 1}{n ^{3}}

n = 1 \sum \infty \frac{7}{6 n}

k = 1 \sum \infty tan (\frac{1}{k})

n = 1 \sum \infty \frac{1}{25 n ^{2} - 1}

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

What is the sum from n=1 to infinity of (ln(2))/(n!) ?
The sum from n=1 to infinity of (ln(2))/(n!) is converges