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Using $$f^n(0)=n! .a_n$$ I have$$\frac{e^x}{1-x}=\sum{\left(\frac{x^n}{n!}\right)}\sum{x^n}=\sum_{n \geq 0}\left({\sum_{k=0}^{n}{\frac{n!}{k!n!}}}\right)$$ How to get the combination?

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  • $\begingroup$ What is $a_n$ ? $\endgroup$ – Andreas Aug 1 '17 at 19:34
  • $\begingroup$ @Andreas $n$th coefficient in Taylor series at $0$ $\endgroup$ – angryavian Aug 1 '17 at 19:35
  • $\begingroup$ This answer seems relevant $\endgroup$ – angryavian Aug 1 '17 at 19:37
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The General Leibniz rule tells us that for two smooth functions $f,g$, the $n$th derivative of $h(x) = f(x)g(x)$ is

$$h^{(n)}(x) = \sum_{k=0}^n {n \choose k} f^{(n-k)}(x)g^{(k)}(x)$$

The $n$ derivative of $e^x$ at $x=0$ is $1$, and the $n$th derivative of $\frac{1}{1-x}$ at $x=0$ is $n!$ by considering the power series. Hence, the $n$th derivative of $h(x) = \frac{e^x}{1-x}$ at $x=0$ is

$$h^{(n)}(x) = \sum_{k=0}^n {n \choose k} (n-k)! = \sum_{k=0}^n \frac{n!}{k!} = n! \sum_{k=0}^n \frac{1}{k!}$$

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Let say you have the product of two power series $\mathrm A(x) \mathrm B(x)$ then you have $$(a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+\cdots)(b_0+b_1x+b_2x^2+b_3x^3+b_4x^4+\cdots)$$

The constant term is given by $a_0b_0$. The $x$-term is $(a_0b_1+a_1b_0)x$. The $x^2$-terms is $$(a_0b_2+a_1b_1+a_2b_0)x^2$$ The $x^3$-term will be $(a_0b_3+a_1b_2+a_2b_1+a_3b_0)x^3$. The $x^k$-term is given by $$(a_0b_k+a_1b_{k-1}+a_2b_{k-2}+\cdots+a_{k-2}b_2+a_{k-1}b_1+a_kb_0)x^k$$ $$=\left( \sum_{i=0}^k a_ib_{k-i}\right)x^k$$

In your case, where $\mathrm A(x) = \mathrm e^x$ and so $a_i = \frac{1}{i!}$, and $\mathrm B(x) = \frac{1}{1-x}$ where $b_j = 1$.

The $x^k$-term in the product of their power series will be $$\left(\sum_{i=0}^k \frac{1}{i!}\cdot 1\right)x^k = \sum_{i=0}^k \frac{x^k}{i!}$$

This gives us the power series. Recall that the Taylor series of a function is $$\sum_{k=0}^{\infty} \frac{\mathrm f^{(k)}(0)}{k!}\,x^k$$

Equating the $x^k$-coefficients gives $$\sum_{i=0}^k \frac{1}{i!}=\frac{\mathrm f^{(k)}(0)}{k!} \ \ \implies \ \ \mathrm f^{(k)}(0) \ \ = \ \ \sum_{i=0}^k \frac{k!}{i!}$$ $$\mathrm f^{(k)}(0) = \frac{k!}{0!} + \frac{k!}{1!} + \frac{k!}{2!} + \cdots + \frac{k!}{(k-2)!} + \frac{k!}{(k-1)!} + \frac{k!}{k!}$$

I'm not able to simplify this any further. Maple suggests the incomplete Gamma function

$$\mathrm e \Gamma(k+1,1) \ \ = \ \ \mathrm e \int_1^{\infty} t^k\mathrm e^{-t}~\mathrm dt$$

I have no idea what this means!

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Let $$y=\frac{e^x}{1-x},$$ or

$$y(1-x)=e^x.$$

By differentiation,

$$y'(1-x)-y=e^x,\\ y''(1-x)-2y'=e^x,\\ y'''(1-x)-3y''=e^x,\\ y''''(1-x)-4y'''=e^x,\\ \cdots$$

and obviously

$$y^{(n)}(1-x)-ny^{(n-1)}=e^x.$$

Evaluating at $0$, we get the recurrence

$$y_n=ny_{n-1}+1$$ with $y_0=1$.


To solve it, let $z_n:=\dfrac{y_n}{n!}$. (By the way, the $z_n$ are precisely the Taylor coefficients.) We have

$$n!z_n=n(n-1)!z_{n-1}+1$$ or

$$z_n=z_{n-1}+\frac1{n!}.$$

Hence,

$$\frac{y_n}{n!}=z_n=\sum_{k=0}^n\frac1{k!}.$$

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  • $\begingroup$ This method is relatively painless as it avoids the product of series. $\endgroup$ – Yves Daoust Aug 1 '17 at 20:12

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