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consider the identity $$\frac{e^{-x}}{1-x}=\sum_{n=0}^{\infty}c_nx^n$$

Show that for each $n\ge0$ $$\sum_{k=0}^{n}\frac{c_k}{(n-k)!}=1$$

My trial : By cauchy product,

$$c_k=\sum_{i=0}^{k}\frac{(-1)^i}{i!}$$

Then $$\sum_{k=0}^{n}\frac{c_k}{(n-k)!}=\sum_{k=0}^{n}\sum_{i=0}^{k}\frac{(-1)^i}{(n-k)!i!}$$ $$=\sum_{i=0}^{n}\sum_{k=i}^{n} \frac{(-1)^i}{(n-k)!i!}$$ $$=\sum_{i=0}^{n}\sum_{k=i}^{n}\binom nk\frac{k!}{n!}$$

I was stuck in here. So, I tried to solve it by taking $c_k=\frac{f^{k}(0)}{k!}$. But I couldn't solve as well.. Could you please give me a few hint.. it will help me a lots. Thanks!

fivestar
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2 Answers2

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Hint: Use induction and Newton's binomial theorem.

Details: $n=0$ clearly satisfies the condition since $c_0=1$. Next assume the formula holds for $n$ and write $$\begin{split} \sum_{k=0}^{n+1}\frac{c_{n+1-k}}{k!} &=\frac{c_0}{(n+1)!}+\sum_{k=0}^n\frac{c_{n-k}+(-1)^{n+1-k}/(n+1-k)!}{k!}\\ &=1+\frac{1}{(n+1)!}\bigg(1+\sum_{k=0}^n\binom{n+1}{k}(-1)^{n+1-k}1^k\bigg) =1+\frac{(1-1)^{n+1}}{(n+1)!}=1, \end{split} $$ concluding the inductive step and the proof.

Jorge I. González C.
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Multiply both sides by $e^x$, on the right-hand side use the Cauchy product, on the left the geometric series and finally compare coefficients of $x^n$ on both sides.

Gary
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