Let $$a = 1 + \frac{^2C_2}{3!} + \frac{^3C_2}{4!} + \frac{^4C_2}{5!} + \ldots$$, $$b = 1 + \frac{^1C_0 + ^1C_1}{1!} + \frac{^2C_0+^2C_1+^2C_2}{2!} + \frac{^3C_0+^3C_1+^3C_2+^3C_3}{3!} + \ldots$$. Then $$\frac{2b}{a^2}$$ is equal to ______.

First write the two series in compact sigma‐notation.

For $$a$$, every term after the first is of the form $$\frac{{}^{n}C_{2}}{(n+1)!}$$ with $$n \ge 2$$.
Hence $$a = 1 + \sum_{n = 2}^{\infty} \frac{{}^{n}C_{2}}{(n+1)!}\; -(1)$$

For $$b$$, the $$n^{\text{th}}$$ bracket gives $$\sum_{r = 0}^{n} {}^{n}C_{r}=2^{n}$$, so

$$b = 1 + \sum_{n = 1}^{\infty} \frac{2^{n}}{n!}\; -(2)$$

Case 1: Evaluation of $$a$$

Use $${}^{n}C_{2} = \frac{n(n-1)}{2}$$ in $$(1):$$

$$a = 1 + \sum_{n=2}^{\infty} \frac{n(n-1)}{2\,(n+1)!}$$

Simplify the general term:

$$\frac{n(n-1)}{(n+1)!} = \frac{n(n-1)}{(n+1)\,n\,(n-1)!} = \frac{1}{(n+1)(n-2)!}$$

Hence

$$a = 1 + \frac12 \sum_{n=2}^{\infty} \frac{1}{(n+1)(n-2)!}$$

Shift the index: put $$k = n-2 \;(k = 0,1,2,\dots)$$.

Then $$a = 1 + \frac12 \sum_{k=0}^{\infty} \frac{1}{(k+3)\,k!}\; -(3)$$

Use the identity $$\frac{1}{k+3} = \int_{0}^{1} x^{k+2}\,dx$$

Insert this into $$(3):$$

$$a = 1 + \frac12 \int_{0}^{1} \left[\sum_{k=0}^{\infty} \frac{x^{k}}{k!}\right] x^{2}\,dx = 1 + \frac12 \int_{0}^{1} x^{2} e^{x}\,dx$$

Integrate $$x^{2}e^{x}$$ once: $$\int x^{2} e^{x} dx = e^{x}(x^{2}-2x+2) + C$$

Therefore $$\int_{0}^{1} x^{2} e^{x} dx = e^{1}(1^{2}-2\cdot1+2) - e^{0}(0-0+2) = e(1) - 2 = e - 2$$

Put this back: $$a = 1 + \frac12 (e - 2) = \frac{e}{2}\; -(4)$$

Case 2: Evaluation of $$b$$

From $$(2)$$ and the Maclaurin series of $$e^{x}$$, $$b = 1 + \sum_{n=1}^{\infty} \frac{2^{n}}{n!} = \sum_{n=0}^{\infty} \frac{2^{n}}{n!} = e^{2}\; -(5)$$

Case 3: Required ratio

Using $$(4)$$ and $$(5):$$

$$\frac{2b}{a^{2}} = \frac{2\,e^{2}}{\left(\dfrac{e}{2}\right)^{2}} = \frac{2\,e^{2}}{e^{2}/4}=8$$

Hence $$\displaystyle \frac{2b}{a^{2}} = 8$$.