Evaluating $\frac{2^1}{1}+\frac{2^2}{1\cdot3}+\frac{2^3}{1\cdot3\cdot5}+\frac{2^4}{1\cdot3\cdot5\cdot7}+\frac{2^5}{1\cdot3\cdot5\cdot7\cdot9}+\cdots$

Question

How do I evaluate this infinite series? $$\frac{2^1}{1}+\frac{2^2}{1\cdot3}+\frac{2^3}{1\cdot3\cdot5}+\frac{2^4}{1\cdot3\cdot5\cdot7}+\frac{2^5}{1\cdot3\cdot5\cdot7\cdot9}+\cdots$$ The approximate value of this series is around $4.05\ldots$

I noticed that each term could be transformed into a product of differences, for instance:

$$\begin{align} \frac{2^3}{1\cdot3\cdot5}&=\left(\frac{1}{1}-\frac{1}{5}\right)\left(\frac{1}{1}-\frac{1}{3}\right) \\[4pt] \frac{2^4}{1\cdot3\cdot5\cdot7}&=\left(\frac{1}{3}-\frac{1}{7}\right)\left(\frac{1}{1}-\frac{1}{5}\right) \\[4pt] &\cdots \end{align}$$

However upon expanding the terms from the resulting sum of products, the fractions do not seem to cancel out.

If you put $x = \sqrt{2}$ in the formula given in the following answer and multiply the final result by $2$, you'll have a closed form (well, closed if you count $\mathrm{erf}$ as an elementary function but still): Sum of $ \frac{1}{1\cdot 3}+\frac{1}{1\cdot 3\cdot 5}+\frac{1}{1\cdot 3\cdot 5\cdot 7}+ \cdots$ - found using Approach$0$ and copy-pasting your formula. — Bruno B, Aug 10 '23 at 08:24
Thank you. I was playing around with the expansion of e and replaced $x^n/n!$ with $x^((2n+1)/2)/Γ(n/2)$. The summation in the question divided by $√π$ was the solution of f(1) — LithiumPoisoning, Aug 10 '23 at 08:52
Here are a few proofs of what is essentially the same formula as the one linked to in Bruno B's comment. — Christian E. Ramirez, Aug 10 '23 at 09:36
I honestly can't tell. I just have a high school education in maths. Making sense of these answers will take some time lol. Regardless thank you — LithiumPoisoning, Aug 10 '23 at 15:19

Evaluating $\frac{2^1}{1}+\frac{2^2}{1\cdot3}+\frac{2^3}{1\cdot3\cdot5}+\frac{2^4}{1\cdot3\cdot5\cdot7}+\frac{2^5}{1\cdot3\cdot5\cdot7\cdot9}+\cdots$

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