Let a=1+frac ^2 C_23 !+frac ^3 C_24 !+frac ^4 C_25 !+...., b=1+frac ^1 C_0+ ^1 C_11 !+frac ^2 C_0+ ^2 C_1+ ^2 C_22 !+frac ^3 C_0+ ^3 C_1+ ^3 C_2+ ^3 C_33 !+.... Then (2 b / a^2) is equal to _________.

a=1+frac ^2 C_23!+frac ^3 C_24!+frac ^4 C_25!+ ... \ b=1+frac ^1 C_0+ ^1 C_11!+frac ^2 C_0+ ^2 C_1+ ^2 C_22!+ ... \ b=1+(2 / 1!)+(2^2 / 2!)+(2 / 3!)+ ... =e^2 Using e^x=1+(x / 1!)+(x^2 / 2!)+(x / 3!)+ ... a= 1+sum _r=2^infinity frac ^r C_2(r+1)!=1+sum _r=2 (r(r-1) / 2(r+1)!) \ =1+(1 / 2) sum _r=2^infinity ((r+1) r-2 r / (r+1)!) \ =1+(1 / 2) sum _r=2^infinity (1 / (r-1)!)-(1 / 2) sum _r=2 (2 r / (r+1)!) \ =1+(1 / 2)((1 / 1!)+(1 / 2!)+ ... )-sum _r=2^infinity ((r+1)-1 / (r+1)!) \ =1+(1 / 2)(e-1)-sum _r=2^infinity (1 / r!)+sum _r=2 (1 / (r+1)!) \ =1+(1 / 2)(e-1)-(e-(1 / 1!)-(1 / 0!)) +(e-(1 / 1!)-(1 / 0!)-(1 / 2!)) \ =1+(e / 2)-(1 / 2)-e+2+e-2-(1 / 2)=(e / 2) \ => (2 b / a^2)=frac2(e^2 / 4) 8 \

Let a1frac 2 C23 frac 3 C24 frac 4 C25 b1frac 1 C0 1 C1... | (Mathematics)

\begin{aligned} & a=1+\frac{{ }^2 C_2}{3!}+\frac{{ }^3 C_2}{4!}+\frac{{ }^4 C_2}{5!}+\ldots \\ & b=1+\frac{{ }^1 C_0+{ }^1 C_1}{1!}+\frac{{ }^2 C_0+{ }^2 C_1+{ }^2 C_2}{2!}+\ldots \\ & b=1+\frac{2}{1!}+\frac{2^2}{2!}+\frac{2}{3!}+\ldots=e^2 \end{aligned}

Using

e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x}{3!}+\ldots

\begin{aligned} a= & 1+\sum_{r=2}^{\infty} \frac{{ }^r C_2}{(r+1)!}=1+\sum_{r=2} \frac{r(r-1)}{2(r+1)!} \\ & =1+\frac{1}{2} \sum_{r=2}^{\infty} \frac{(r+1) r-2 r}{(r+1)!} \\ & =1+\frac{1}{2} \sum_{r=2}^{\infty} \frac{1}{(r-1)!}-\frac{1}{2} \sum_{r=2} \frac{2 r}{(r+1)!} \\ & =1+\frac{1}{2}\left(\frac{1}{1!}+\frac{1}{2!}+\ldots\right)-\sum_{r=2}^{\infty} \frac{(r+1)-1}{(r+1)!} \\ & =1+\frac{1}{2}(e-1)-\sum_{r=2}^{\infty} \frac{1}{r!}+\sum_{r=2} \frac{1}{(r+1)!} \\ & =1+\frac{1}{2}(e-1)-\left(e-\frac{1}{1!}-\frac{1}{0!}\right) \\+\left(e-\frac{1}{1!}-\frac{1}{0!}-\frac{1}{2!}\right) \\ & =1+\frac{e}{2}-\frac{1}{2}-e+2+e-2-\frac{1}{2}=\frac{e}{2} \\ \Rightarrow & \frac{2 b}{a^2}=\frac{2}{\frac{e^2}{4}} 8 \end{aligned}

Explanation

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