Question

If $e^x/1-x=B_0+B_{1}x+B_2{x}^2+....+B_n{x}^n+...,then{B}_n-B_{n-1}$equals

Answer 1

Finding the value of $\mathbf{}{\mathit{B}}_{\mathbf{n}}\mathbf{-}{\mathit{B}}_{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{}$:

Given that $e^x/1-x=B_0+B_{1}x+B_2{x}^2+....+B_n{x}^n+...,$

$$\begin{gathered} e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+....\infty \\ {(1-x)}^{-1}=1+x+x^2+...+\infty \\ {e}^x/(1-x)=e^x{(1-x)}^{-1} \\ =\{1+\frac{x}{1!}+\frac{x^2}{2!}+....\infty \}\{1+x+x^2+...+\infty \}=B_0+B_{1}x+B_2{x}^2+....+B_n{x}^n+\{1+\frac{x}{1!}+\frac{x^2}{2!}+....\infty \}+x\{1+\frac{x}{1!}+\frac{x^2}{2!}+....\infty \}+x^2\{1+\frac{x}{1!}+\frac{x^2}{2!}+....\infty \}... \\ =B_0+B_{1}x+B_2{x}^2+....+B_n{x}^n+... \\ =B_0=1,B_1=1+\frac{1}{1!},B_2=1+\frac{1}{1!}+\frac{1}{2!}, \\ {B}_n=1+\frac{1}{1!}+\frac{1}{2!}+......+\frac{1}{n!} \\ {B}_n-B_{n-1}=1+\frac{1}{1!}+\frac{1}{2!}+......+\frac{1}{n!}-\{1+\frac{1}{1!}+\frac{1}{2!}+......+\frac{1}{(n-1)!}\} \\ =\frac{1}{n!} \end{gathered}$$Hence, the value of $\mathbf{}{\mathit{B}}_{\mathbf{n}}\mathbf{-}{\mathit{B}}_{\mathbf{n}\mathbf{-}\mathbf{1}}$ is $\frac{\mathbf{1}}{\mathbf{n}\mathbf{!}}$.