Measure of a modified Cantor set Suppose a modified Cantor set:
Starting with $E_0 = [0,1]$, we delete the middle interval of length $1/3$, then we delete the middle intervals of length $1/15$, and so on; in each step we delete from all the intervals of set $E_m$ the middle of length 
$$\frac{1}{3\cdot5\cdot\ \space ... \space\cdot(2m+1) \cdot (2m+3)},$$ and we have to prove that the measure of the interesection is between $0$ and $1$. 
I found the total length that we toss away but I am not able to bound the sum of the series. Any ideas? The series is actually: 
$$\sum^\infty_{m=0}\frac{2^m}{3\cdot5\cdot\ \space ... \space\cdot(2m+1) \cdot (2m+3)}$$
Thank you in advance!

How about this:
$$\sum_{n=0}^\infty \frac{2^n}{3 \cdot 5 \cdot \ldots \cdot (2n+3)} \ge 1/ 3 $$
For the opposite direction:
$$\sum_{n=0}^\infty \frac{2^n}{3 \cdot 5 \cdot \ldots \cdot (2n+3)} \le  \sum_{n=0}^\infty \frac{2^n}{2 \cdot 4 \cdot \ldots \cdot (2n+2)} = \sum_{n=0}^\infty \frac{2^n}{2^{n+1} (n+1)!} = \frac 1 2 \sum_{n=0}^\infty \frac{1}{(n+1)!} \le \frac{e-1}{2} \approx 0.859141$$
