The sum of finitely many geometric sequences has a limit iff every base is less than 1 in absolute value 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?
 A: 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.
A: 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.
