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I have a question of proving or disproving a specific sum converges to $e^{A\cdot B(t)}$. $t$ is a parameter.

My sum is something like: $\sum_{k=0}^\infty \frac{1}{k!} f(k,t)$ .

The question is, if I present the exponential as: $e^{A\cdot B(t)}=\sum_{k=0}^\infty \frac{1}{k!} (A\cdot B(t))^k$, can I say that proving that no such $f(k,t)$ exists such that $f(k,t)=(A\cdot B(t))^k$ is to prove that they are necessarily not equal?

At first I thought that the answer is no, since a sum can be presented in more ways then one, but then I though that at any rate, if such a sum exists, it has a taylor series representation as I specified above... I'm puzzled.

Thanks.

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