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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 $\lambda$.

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    Isn't there a condition that it should be true for any $\lambda$? Besides, I don't see the point of having $n\lambda$. Just call it $x$. Unless there is something you are not telling us here, but then you should. If it's just for one value of $\lambda$ and $n$, it's easy to construct a counter example. (Think Taylor expansion of $\sin$ or $\cos$.)2011-03-20
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    $n$ doesn't seem relevant to the question. I think you are missing a quantifier here: don't you want the identity to hold for all $\lambda$?2011-03-20
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    Thanks @Raskolnikov and @Qiaochu Yuan, I've edited the question.2011-03-20

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