Sum of reciprocal products

$\begin{aligned}\text{The n th term} &\text{ of the sequence is given by}\\ T_{n}&=\dfrac{2n-1}{2^n n!}\\ S&=\sum_{n=1}^{\infty}T_{n}\\ &=\sum_{n=1}^{\infty}\dfrac{2n-1}{2^n n!}\\ &=\sum_{n=1}^{\infty}\dfrac{1}{2^{n-1} (n-1)!}-\dfrac{1}{2^n n!}\\ \text{We can easily} &\text{ see that the sum telescopes,}\\ S&=\dfrac{1}{2^{1-1} (1-1)!}\\ &=\color{#EC7300}\boxed{\color{#333333}1}\end{aligned}$

Observation of this sum tells us all numerators are odd. Wasn't that written as $2k-1$ ? It actually is and the denominators are all products of consecutive even numbers which seems to be the same as $(2k)!!$ , where the double exclamation mark stands for ‘double factorial’ or ‘semifactorial’ and can be rewritten as a single factorial. So now let us rewrite the different terms.

$\begin{aligned} S &= \dfrac{2\cdot \red{1}-1}{\red{1}\cdot 2} + \dfrac{2\cdot \red{2}-1}{\red{2}\cdot 2\cdot \red{1}\cdot 2} + \dfrac{2\cdot \red{3}-1}{\red{3}\cdot 2\cdot \red{2}\cdot 2\cdot \red{1}\cdot 2} + \dfrac{2\cdot \red{4}-1}{\red{4}\cdot 2\cdot \red{3}\cdot 2\cdot \red{2}\cdot 2\cdot \red{1}\cdot 2)} + \ldots\\[18pt] &= \sum_{k=1}^{\infty}\frac{2k-1}{2^k k!}\\[18pt] &= \sum_{k=1}^{\infty}\frac{2k}{2^kk!}\red{\cdot\frac{k^{-1}}{k^{-1}}} - \frac{1}{2^kk!}\\[18pt] &= \sum_{k=1}^{\infty}\frac{2}{2^k(k-1)!} - \frac{1}{2^kk!}\\[18pt] &= \sum_{k=1}^{\infty}\frac{1}{2^{k-1}(k-1)!} - \frac{1}{2^kk!} \enspace\red{\text{(‘telescoping sum’: most terms will cancel out)}}\\[18pt] &= \frac11 \red{- \frac12 + \frac12 - \frac18 + \frac18 - \frac{1}{48} + \frac{1}{48} - \frac{1}{384} + \frac{1}{384} -+ \ldots} - \frac{1}{\infty}\\[18pt] &= 1 - 0 = \boxed{1} \end{aligned}$

Just for the sake of diversity, a solution with Taylor's series:

re-write the original problem

$\displaystyle \frac{1}{2}+\frac{3}{4\times 2}+\frac{5}{6\times 4\times 2}+\cdots=\frac{1}{2}+\frac{3}{2^2\cdot 2!}+\frac{5}{2^3\cdot 3!}+\cdots$

We know that

$\displaystyle e^{x^2}-1= x^2+\frac{x^4}{2!}+\frac{x^6}{3!}+\cdots$

$\displaystyle \frac{e^{x^2}-1}{x}= x+\frac{x^3}{2!}+\frac{x^5}{3!}+\cdots$

differentiate both sides:

$\displaystyle \frac{(2x^2-1)e^{x^2}+1}{x^2}= 1+\frac{3x^2}{2!}+\frac{5x^4}{3!}+\cdots$

$\displaystyle \frac{(2x^2-1)e^{x^2}+1}{2x^2}= \frac{1}{2}+\frac{3x^2}{2\cdot 2!}+\frac{5x^4}{2\cdot 3!}+\cdots$

now plug in $\displaystyle x=\frac{1}{\sqrt{2}}$

$\displaystyle 1 = \frac{1}{2}+\frac{3}{2^2\cdot 2!}+\frac{5}{2^3\cdot 3!}+\cdots$

Use harmonic progression with telescoping series

most simple answer:

$\frac{1}{2} + \frac{3}{4\times2} + \frac{5}{6\times4\times2} + \frac{7}{8\times6\times4\times2} + \cdots$

converting the numerator to a subtraction:

$\frac{1}{2} + \frac{4-1}{4\times2} + \frac{6-1}{6\times4\times2} + \frac{8-1}{8\times6\times4\times2} + \cdots$ (the key is see that the substraction in the numerator has the max value of the denominator and 1 xD)

splitting fractions and simplifying factors:

$\frac{1}{2} + (\frac{4}{4\times 2} - \frac{1}{4\times 2}) + (\frac{6}{6\times4\times2} - \frac{1}{6\times4\times2}) + (\frac{6}{8\times6\times4\times2} - \frac{1}{8\times6\times4\times2}) + \cdots$

$\frac{1}{2} + (\frac{1}{2} - \frac{1}{4\times 2}) + (\frac{1}{4\times2} - \frac{1}{6\times4\times2}) + (\frac{1}{6\times4\times2} - \frac{1}{8\times6\times4\times2}) + \cdots$

cancelling similar terms:

$\frac{1}{2} + \frac{1}{2} + 0 + 0 + 0 + \cdots = 1$