Series Sum 5C1Ad1

1. **Problem statement:** Find the sum of the series $$1 + \frac{3}{1!} + \frac{5}{2!} + \frac{7}{3!} + \cdots$$ 2. **General term:** Let the $r^{th}$ term be $$U_r = \frac{2r-1}{(r-1)!}$$ for $r \geq 1$. 3. **Rewrite the numerator:** $$U_r = \frac{2r-2+1}{(r-1)!} = \frac{2(r-1)}{(r-1)!} + \frac{1}{(r-1)!} = \frac{2}{(r-2)!} + \frac{1}{(r-1)!}$$ 4. **Sum of first $n$ terms:** $$S_n = \sum_{r=1}^n U_r = \sum_{r=1}^n \left( \frac{2}{(r-2)!} + \frac{1}{(r-1)!} \right) = 2 \sum_{r=1}^n \frac{1}{(r-2)!} + \sum_{r=1}^n \frac{1}{(r-1)!}$$ 5. **Index shift for sums:** - For the first sum, let $k = r-2$, then when $r=1$, $k=-1$ (term undefined), so start from $r=2$: $$2 \sum_{r=2}^n \frac{1}{(r-2)!} = 2 \sum_{k=0}^{n-2} \frac{1}{k!}$$ - For the second sum, let $m = r-1$, then when $r=1$, $m=0$: $$\sum_{r=1}^n \frac{1}{(r-1)!} = \sum_{m=0}^{n-1} \frac{1}{m!}$$ 6. **As $n \to \infty$, these sums approach the exponential series:** $$\sum_{k=0}^\infty \frac{1}{k!} = e$$ 7. **Therefore, the infinite sum is:** $$S = 2e + e = 3e$$ **Final answer:** $$\boxed{3e}$$ **Conclusion:** The solution and answer given are correct.