I wondered how continually compounding a fraction in the denominator would behave as the successive denominators increment. What I was doing was essentially:
$\dfrac 1 {\dfrac 2 {\dfrac 3 {\dfrac 4 {\dfrac 5 {\dfrac 6 {\dfrac 7 {\dfrac 8 {\dfrac 9 {\vdots}}}}}}}}}$
I noticed that the result oscillated between numbers that were above 1 and below 1, and I realized that the pattern was actually progressing with respect to adding two numbers, i.e. the series is 1/2, 1/(2/(3/4)), etc.
Following this trail it appears that the limit as the number of (pairs of) denominators we add approaches infinity, this number approaches zero, at least as far as I can tell.
Consulting OEIS (A001147, A000165) resulted in my inspiration to find the limit of the infinite sum as we travel down the fraction, and I was able to create what I believe is the correct formula:
$\displaystyle S_\infty=\sum_{i=1}^\infty{\prod_{n=1}^i{\dfrac{2n-1}{2n}}}$
So, what I am wondering is,
- what is the solution to this, assuming it converges?
- how can I (properly) prove that the expression above is also equal to the expression below (I am confident it is)?
$\displaystyle \sum_{n=1}^\infty{\dfrac{1}{4^n} \binom{2n}n}$
Personally hoping it's convergent, since I believe the terms tend to zero.
Consider $$ S_p=\sum_{i=1}^p{\prod_{n=1}^i{\dfrac{2n-1}{2n}}}=\sum_{i=1}^p \frac{\Gamma \left(i+\frac{1}{2}\right)}{\sqrt{\pi }\,\Gamma (i+1)}=\frac{2 \,\Gamma \left(p+\frac{3}{2}\right)}{\sqrt{\pi }\, \Gamma (p+1)}-1$$ Using Stirling approximation $$S_p=-1+2\sqrt{\frac p \pi}\left(1+\frac{3}{8 p}-\frac{7}{128 p^2}+O\left(\frac{1}{p^3}\right)\right)$$ On the other hand $$T_p=\displaystyle \sum_{n=1}^p{\dfrac{1}{4^n} \binom{2n}n}=-1+2^{-2 p-1} (p+1) \binom{2p+2}{p+1}=\frac{2 \Gamma \left(p+\frac{3}{2}\right)}{\sqrt{\pi } \Gamma (p+1)}-1$$