$\sum_{k = 0}^{\infty} a_{k}\frac{\lambda^{k}}{k!} = 0$ for all $\lambda > 0 \implies a_{k} = 0 \forall k$

Question

Let $a_{k} \in \mathbb{R}$ for each $k \in \mathbb{N}$

I'm trying to prove that if $\sum_{k = 0}^{\infty} a_{k}\frac{\lambda^{k}}{k!} = 0$ for all $\lambda > 0 \implies a_{k} = 0 \forall k \in \mathbb{N}$

My attempt:

If all $a_{k}\geq0$ or all $a_{k}\leq0$ then it's holds

But when I try to make the general case I get stuck

If $a_{k} \not= 0$ for some $k \in \mathbb{N}$

Let $m = min\{k \in \mathbb{N} : a_{k} \not= 0\}$

Then $-a_{m}\frac{\lambda^{m}}{m!} = \sum_{k=m+1}^{\infty}a_{k}\frac{\lambda^{k}}{k!}$ for all $\lambda >0$

Supposed $a_{m} > 0$

Then $-\infty =\lim_{\lambda \to \infty} -a_{m}\frac{\lambda^{m}}{m!} = \lim_{\lambda \to \infty} \sum_{k=m+1}^{\infty}a_{k}\frac{\lambda^{k}}{k!}$

And I don't know how to continue

Answer 1

Differentiate! $$a_k=\Big(\frac{d}{d\lambda}\Big)^k\sum\cdots\big|_{\lambda=0}=0$$