I'm trying to find some function $g(k)$ such that $$\sum_{k=0}^{\infty} g(k) \frac{(n \lambda)^k}{k!} = 0 $$ The textbook says that there is only one solution, that is $g(k)=0$ for all $k$. But I cannot see why it is so. It is also constrained that $g(k)$ depends on $k$ alone and does not depend on $\lambda$ or $n$. $n$ is a positive integer, $\lambda$ is a positive real. My intuitive feeling is that $g(k)$ can take alternating positive or negative values such that the summation is zero, but I cannot prove how. Any ideas ? Or is $g(k)=0$ the only solution ?
EDIT: The above must hold true for all $\lambda$.
a convergent power series of the form $$f(x)=\sum_{k=0}^{\infty}\frac{g(k)}{k!}x^k$$ represents the zero function iff $g(k)=0$ for all $k$ (here $g(k)=f^{(k)}(0)$). this can be zero for given values of $x$ as noted in the comments e.g. $$ \sin(\pi/2)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}(\pi/2)^{(2k+1)} $$ corresponding to a not-identically-zero $g(k)$ and $\lambda n=\pi/2$.