1. ## Evaluate this sum

Is there any way to evaluate this sum exactly?

$\frac{2}{3^1}+\frac{2}{3^2}+\frac{2}{3^6}+\frac{2} {3^{24}}+... = \sum_{n=1}^{\infty}\frac{2}{3^{n!}}$

If not, can someone give me a good approximation? Also, is there any way to know whether this is rational or irrational? How about algebraic or transcendental?

2. Originally Posted by redsoxfan325
Is there any way to evaluate this sum exactly?

$\frac{2}{3^1}+\frac{2}{3^2}+\frac{2}{3^6}+\frac{2} {3^{24}}+... = \sum_{n=1}^{\infty}\frac{2}{3^{n!}}$

If not, can someone give me a good approximation? Also, is there any way to know whether this is rational or irrational? How about algebraic or transcendental?
let $S=\sum_{k=1}^{\infty} \frac{1}{3^{k!}}.$ for a fixed $n$ let $S_n=\sum_{k=1}^n \frac{1}{3^{k!}}=\frac{p}{q}.$ then: $\left |S-\frac{p}{q} \right|=\sum_{k=n+1}^{\infty} \frac{1}{3^{k!}}< \sum_{k=(n+1)!}^{\infty} \frac{1}{3^k}=\frac{1}{3^{(n+1)! - 1}} \leq \frac{1}{q^n},$ where $q=3^{n!}.$ thus $S$ is a Liouville number and so it's transcendental.

3. Hello,

I don't believe it's possible to find the exact value, but a good approximation : $\color{red}\frac{65}{729}$