Can the sum of two divergent geometric series converge such that none of the terms is zero? So yes i was solving an exercise and i want to split the sum. And i was questioning myself if i can do that ?
for example
$\sum_{i=0}^{\infty}(2^i-1)x^i$ and splitting
$\sum_{i=0}^{\infty}(2x)^i$$-$$\sum_{i=0}^{\infty}x^i$
and saying now that it converges for every $|x|<1/2$ What would be incorrect about this or how could i argument if it was correct ?
 A: I see there is a bounty for this question, but what question? The title says "Can the sum of two divergent geometric series converge such that none of the terms is zero?" and I see that it was already answered by Chris Sanders. In the text there is perhaps a more general question: "...  I want to split the sum. And i was questioning myself if i can do that ?".  A question regarding series in general, not just geometric series.
In general, the sum of two convergent series is a convergent series. If 
$$\sum_{n=1}^{\infty} a_n = A $$
and 
$$ \sum_{n=1}^{\infty} b_n = B$$
then
$$\sum_{n=1}^{\infty} \left(a_n + b_n \right) = \sum_{n=1}^{\infty} a_n + \sum_{n=1}^{\infty} b_n = A+B$$
If one of the two series $\sum_{n=1}^{\infty} a_n $ and $\sum_{n=1}^{\infty} b_n $  is convergent and the other is not convergent, $\sum_{n=1}^{\infty} \left(a_n + b_n \right)$ is not convergent.
If both series $\sum_{n=1}^{\infty} a_n $ and $\sum_{n=1}^{\infty} b_n $  are not convergent, nothing can be said about $\sum_{n=1}^{\infty} \left(a_n + b_n \right)$
In your exercise, both power series $\sum_{i=0}^{\infty}(2x)^i$ and   $\sum_{i=0}^{\infty}x^i$ are convergent for  $|x|<1/2$, so
also $\sum_{i=0}^{\infty}(2^i-1)x^i=\sum_{i=0}^{\infty}(2x)^i-\sum_{i=0}^{\infty}x^i$ is convergent for  $|x|<1/2$. 
For  $1/2<|x|<1$,  $\sum_{i=0}^{\infty}x^i$ is convergent, but $\sum_{i=0}^{\infty}(2x)^i$ is not, so $\sum_{i=0}^{\infty}(2^i-1)x^i$ is not convergent.
For $|x| > 1$, both power series $\sum_{i=0}^{\infty}(2x)^i$ and   $\sum_{i=0}^{\infty}x^i$ are not convergent, but this does not imply that $\sum_{i=0}^{\infty}(2^i-1)x^i$ is not convergent. 
However we know that the set of convergence of a power series is an interval  , therefore, if there were some $x_0$ such that $|x_0| > 1$ and $\sum_{i=0}^{\infty}(2^i-1)x_0^i$ is convergent, then the series would be convergent for $|x| < |x_0|$, so it would be convergent also for $1/2<|x|<1$, but this contradicts what we proved before. The power series $\sum_{i=0}^{\infty}(2^i-1)x^i$ is not convergent for $|x| > 1$. 
Conclusion: the power series $\sum_{i=0}^{\infty}(2^i-1)x^i$ is convergent for  $|x|<1/2$.   
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A: Show that the power series $\sum_{i=0}^{\infty}(2^i-1)x^i$ has convergence radius $1/2$. For $|x|<1/2$ we then have
$\sum_{i=0}^{\infty}(2^i-1)x^i=\sum_{i=0}^{\infty}(2x)^i-\sum_{i=0}^{\infty}x^i$ ,
since both power series on the right are convergent for  $|x|<1/2$.
A: I will answer the question in the title. 
Can $\sum_{i=0}^{\infty}p^i-q^i$ converge, if $p\neq q$ and $|p|,|q|\geq1$? No. 
This is obvious if $p=-q$. Without loss of generality, assume $|p|>|q|$. Then $p^n-q^n$ will eventually exceed $1$ (prove it).
