# Generating function for a sequence

Please provide a clue on how to solve the following problem:

Find a closed form for the generating function for the sequence $\{a_n\}$, where $$a_n = \frac1{(n+1)!}$$ for $n=0,1,2,\ldots$

I know this looks like $$e^x = \frac{1}{k!} = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$ and I can start from this but I do not know how to proceed.

Is the following procedure valid? If not, what is wrong and why?

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

So, finally we have

$$e^x = 1 + \frac{1}{(k+1)!}$$

Therefore, $$\frac{1}{(k+1)!} = e^x - 1$$

I will very much appreciate your feedback.

• Any comment about the above procedure? – JORGE Apr 13 '15 at 14:13
• You write all sorts of equalities that make no sense. You have to be careful to distinguish between the sequence $\{a_n\}$, a particular element $a_k$, and its generating function $A(z) = \sum_{n \ge 0} a_n z^n$. Often one writes the relation between the sequence and the generating function $\{a_n\} \leftrightarrow A(z)$. – vonbrand Jul 25 '15 at 13:39

Hint: $\displaystyle\sum_{n\ge0}a_nx^n=\frac1x\cdot\sum_{n\ge0}\frac{x^{n+1}}{(n+1)!}=\frac1x\cdot\sum_{n\ge1}\frac{x^n}{n!}$.

$$e=\sum_{n=0}^{\infty }\frac{1}{n!}=1+\sum_{n=1}^{\infty }\frac{1}{(n)!}=1+\sum_{n=0}^{\infty }\frac{1}{(n+1)!}$$

hence $$\sum_{n=0}^{\infty }\frac{1}{(n+1)!}=e-1$$

• I am sorry, I deleted the comment accidentally. I want to say that your solution looks very logic so I edited my original question and included the complete process based on your suggestion. However, I found the following solution $\frac{e^x - 1}{x}$ in a book of answers. – JORGE Apr 12 '15 at 14:12
• That is the sum of the secuence, not it's generating function. – vonbrand Jul 25 '15 at 13:35

Cleaned up version of the question, with correct answer: Find the generating function for the sequence $\{1 / (n + 1)!\}$.

The generating function is:

\begin{align} G(z) &= \sum_{n \ge 0} \frac{1}{(n + 1)!} z^n \\ &= \frac{1}{z} \sum_{n \ge 0} \frac{z^{n + 1}}{(n + 1)!} \\ &= \frac{1}{z} \left( \sum_{n \ge 0} \frac{z^n}{n!} - 1 \right) \\ &= \frac{\mathrm{e}^z - 1}{z} \end{align}