How to estimate $\sum_{n=0}^\infty (\log\log2)^n/n! $ from below?

Question

How to show that the following inequality holds:

$$\sum_{n=0}^\infty \frac{(\log\log2)^n}{n!}>\frac 35$$

Is it possible to prove this using induction?

Answer 1

Hint: what is $\sum_n \frac{x^n}{n!}$? Then plug $x=\log\log2$.

Answer 2

Even if you don't know what $\sum\frac{x^n}{n!}$ is, the first two terms give $1+\log\log2\approx0.63\ldots>\frac35$. Since this is an alternating series and $1+\log\log2$ is a partial sum ending on a negative term, whatever the sum is it will be greater than that $0.63\ldots$.