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It is well known that a geometric sequence $x_n = c b^n$ with a base $b\ne 1$ has a limit if and only if $|b|<1$. (Note that $b$ might be complex.)

I'd like to prove a more general fact: a finite sum of such sequences, with distinct bases $b_1, \dots, b_r$, not equal to $1$, has a limit if and only if $|b_1|, \dots, |b_r|<1$. More precisely: if $$x_n = c_1 b_1^n + c_2 b_2^n + \dots + c_r b_r^n$$ where $r$ is a fixed integer, $c_k$ are fixed nonzero complex numbers, and $b_k$ are fixed distinct complex numbers not equal to $1$, then $\lim_{n\to\infty} x_n$ exists if and only if $\max_k |b_k|<1$.

Progress. Sufficiency is obvious, the question is necessity. The terms with larger modulus dominate the rest, thus we may assume $|b_1|=\dots = |b_r|$. Also, it suffices to consider the case $|b_1|=\dots = |b_r| = 1$ because if the common modulus is $M>1$ and the limit exists, then multiplying the sequence by $M^{-n}$ will result in another convergent sequence.

So, $b_k = \exp(2\pi i \theta_k)$ for some $\theta_k\in (0, 1)$. If all $\theta_k$ are rational, the sequence $x_n$ is periodic (and nonconstant), so there is no limit. How to deal with the case when some $\theta_k$ are irrational?

  • If $\forall k \in \mathbb{N}, \theta_k = 0 \implies b_k = 1 \implies x_n = \sum_{i=1}^r c_i = C$. Thus, $\lim_{n \to +\infty} x_n = C$. It is not clear what you're asking here. Can you detail why you think $x_n$ is periodic? – Raito Feb 08 '18 at 07:12
  • The bases are assumed not equal to 1. –  Feb 08 '18 at 07:25
  • I don't get it, if $$x_n = -2 \cdot 3^n + 2 \cdot 3^n,$$ then does it have limit? – dEmigOd Feb 08 '18 at 09:02
  • The bases are assumed to be distinct. –  Feb 08 '18 at 15:38

2 Answers2

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You are right, and this is easy to show. One has the identity

$$ \left(\begin{array}{c} x_n \\ x_{n+1} \\ \vdots \\ x_{n+r-1} \\ \end{array}\right) = \left(\begin{array}{cccc} c_1 & c_2 & \ldots & c_r \\ b_1c_1 & b_2c_2 & \ldots & b_rc_r \\ \vdots & \vdots & \ldots & \vdots \\ b_1^{r-1}c_1 & b_2^{r-1}c_2 & \ldots & b_r^{r-1}c_r \\ \end{array}\right) \left(\begin{array}{c} b_1^n \\ b_2^n \\ \vdots \\ b_r^n \\ \end{array}\right) $$

Note that the matrix above is a Vandermonde matrix, so it is invertible because the $b_i$ are distinct. It follows that $(b_1^n)$ is a linear combination of $x_n,x_{n+1},\ldots,x_{n+{r-1}}$. So if $(x_n)$ converges to some $x_{\infty}$, so do $,x_{n+1},\ldots,x_{n+{r-1}}$, and $(b_1^n)$ therefore has a limit also, which is known to be false.

Ewan Delanoy
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Let us consider only the case you stated as remaining. The fact that the $b_k$ are rational or irrational will not be important.
Let us prove by induction on $r$ . The base case is the "well known" fact you mentioned.
Now suppose (in order to get an absurd) that there is such a converging sequence for $r$ terms but none for fewer than $r$ terms.
Let us call this sequence $x_n$ and define the sequence $$y_n = x_{n+1} - b_1 x_n$$ As $x_n$ converges , it follows that $y_n$ converges as well, for it is a linear combination of converging sequences. However $y_n$ can be written as : $$y_n = c_2 (b_2 - b_1) b_2^n + \dots + c_r (b_r - b_1) b_r^n$$ This sequence has $r-1$ terms, the bases have ,modulus 1 are are distinct, and the coefficients are nonzero, for $ c_k (b_k - b_1)$ is nonzero as neither $c_k$ nor $(b_k - b_1)$ are zero , as the bases are all distinct.
This shows a contradiction that proves the induction.

user2759511
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