Question

Find the sequence with each of these functions as its exponential generating function $f\left(x\right)=e^x-\left(\frac{1}{1+x}\right).$

Short answer

$a_k=1-{(-1)}^{k}k!$

Step-by-step solution

Given data

The given function is $f\left(x\right)=e^x-\left(\frac{1}{1+x}\right)$.

Concept used of generating function

The ordinary generating function of a sequence ${\mathit{a}}_{\mathbf{n}}$is

$G\left(a_n;x\right)=\sum _{n=0}^{\infty }{a}_n{x}^n.$

Solve the function

Using $\sum _{k=0}^{+\infty }\frac{a_k}{k!}{x}^k=e^x$, we have

$e^x-\frac{1}{1+x}=\sum _{k=0}^{+\infty }\frac{x^k}{k!}-\frac{1}{1-(-x)}$

$=\sum _{k=0}^{+\infty }{x}^k\frac{x^k}{k!}-\sum _{k=0}^{+\infty }{(-x)}^k$

$=\sum _{k=0}^{+\infty }{x}^k\frac{x^k}{k!}-\sum _{k=0}^{+\infty }{(-1)}^{k}k!\frac{x^k}{k!}$

$=\sum _{k=0}^{+\infty }\left(1-{(-1)}^{k}k!\right)\frac{x^k}{k!}$

The sequence $\left\{a_k\right\}$are then the coefficients of $\frac{x^k}{k!}$in the above sum: $a_k=1-{(-1)}^{k}k$.!.