Series of powers Consider the sequences $a_k,b_k\in\mathbb R$ and let $N\in\mathbb N$. Do series of the form
$$f(n)=\sum_{k=1}^N a_k\,{b_k}^n$$
have a name? They seem quite a natural object to study but I'm not sure what to search for online and I'm not coming across them.
I want to prove the following fact about them: if two series of the above form, $f(n)=
\sum_{k=1}^{N} a_k\,{b_k}^n$ and $g(n)=
\sum_{k=1}^{M} c_k\,{d_k}^n$ satisfy $f(n)=g(n)$ for all $n\in\mathbb N$, then $N=M$, $a_k=c_k$ and $b_k=d_k$ for all $1\leqslant k\leqslant N$.

Here is my idea: if $f(n)=g(n)$ for all $n$, then $f(n)/g(n)\to 1$, but the dominant term $a_r{b_r}^n$ in the numerator and dominant term $c_s{d_s}^n$ in the denominator will cause $f(n)/g(n)$ to diverge unless $b_r=d_s$, in which case the limit is $a_r/c_s$, but since the limit is $1$ this forces $a_r=c_s$ too.
If we subtract the dominant terms off, considering $[f(n)-a_r{b_r}^n]/[g(n)-a_r{b_r}^n]$, the same argument establishes the analogous fact for the next pair of dominant terms, and so on, until we either run out of terms from either the numerator and the denominator. But if we run out, then we get a limit of the form $\frac0\infty$ or $\frac\infty0$, which contradicts that $f$ and $g$ coincide.
Is there a nicer way to prove this?
 A: So here is an answer with some extra conditions which we worked out in the comments of the counter-example to the original question.
Let
$$\begin{align}
f(n) &= \sum_{k=1}^N a_k b_k^n \\
g(n) &= \sum_{k=1}^M c_k d_k^n
\end{align}$$
be two functions with $N$-tuples $(a_k)$, $(b_k)$ and $M$-tuples $(c_k)$, $(d_k)$ of non-zero real (or complex) numbers such that:
$$|b_k| > |b_{k+1}|\qquad\text{and}\qquad |d_k| > |d_{k+1}|.\tag 1$$
Then we have
$$ f(n) = g(n) \ \forall n\in\Bbb N \qquad\implies\qquad  M=N, a_k=c_k, b_k=d_k.$$
Proof: We have
$$\begin{align}
f(n) &= \sum_{k=1}^N a_k b_k^n \\
&= a_1b_1^n + \sum_{k=2}^N a_k b_k^n \\
&= b_1^n \left(a_1 + \sum_{k=2}^N a_k (b_k/b_1)^n \right)\\
&= d_1^n \left(c_1 + \sum_{k=2}^M c_k (d_k/d_1)^n \right) = g(n)\\
\end{align}$$
The remaining sums starting at $k=2$ converge to zero for $n\to\infty$ due to $(1)$, or are already zero if a sum is empty. Thus we can divide both sides provided $n$ is large enough:
$$
\left(\frac{b_1}{d_1}\right)^{\!n}
= \frac{c_1 + \sum_{k=2}^M c_k (d_k/d_1)^n}{a_1 + \sum_{k=2}^N a_k (b_k/b_1)^n}
\quad\stackrel{\text{as }n\to\infty}\longrightarrow \quad \frac{c_1}{a_1} \notin\{0,\infty\}
$$
As the limit of the right side exists and is non-zero, it must also exist for the left side.  The only way that $(b_1/d_1)^n$ converges to a finite, non-zero value is to have $b_1=d_1$ and hence also that $a_1=c_1$.
The proof is then finished by repeating the same argument where the 1st summand of either function is dropped because they equal each other, so we are left with sums that are 1 term shorter.
One residual case is remaining if $N\neq M$ if one sum is exhausted (empty) and the other is not. I'll leave that as an exercise.
A: That's not correct? Take for example:
Edit2: So you adjusted the conditions again so that no zeros are allowed in any of $a_k$ or $b_k$.  As an updated counter-example, take $N=M$ even and:
$$\begin{align}
a &= a_1, a_2, a_3, a_4, \dots, a_{N-1}, a_N\\
b &= b_1, b_2, b_3, b_4, \dots, b_{N-1}, b_N\\
c &= a_2, a_1, a_4, a_3, \dots, a_N, a_{N-1}\\
d &= b_2, b_1, b_4, b_3, \dots, b_N, b_{N-1}\\
\end{align}$$
and such that $a_k > 0$, $b_k > 0$ and

*

*$a_k\neq a_{k+1}$ for any odd $k$, and

*$b_k\neq b_{k+1}$ for any odd $k$.

This means you can just swap values at odd and even indices.
Actually, you can apply any non-trivial permutation $\sigma\in{\cal S}_N$ to get a similar counter example, in particular you can pick a permutation that has no fixed-points.
The only case when there is no non-trivial permutation is the case $N=M=1$ which is not very interesting.
