A different notion of convergence for this sequence? I was thinking about sequences, and my mind came to one defined like this:
-1, 1, -1, 1, 1, -1, 1, 1, 1, -1, 1, 1, 1, 1, ...
Where the first term is -1, and after the nth occurrence of -1 in the sequence, the next n terms of the sequence are 1, followed by -1, and so on. Which led me to perhaps a stronger example, 
-1, 1, -1, 1, 1, -1, 1, 1, 1, 1, -1, 1, 1, 1, 1, 1, 1, 1, 1, -1, ...
Where the first term is -1, and after the nth occurrence of -1 in the sequence, the next $2^n$ terms of the sequence are 1, followed by -1, and so on.
By the definition of convergence or by Cauchy's criterion, the sequence does not converge, as any N one may choose to define will have an occurrence of -1 after it, which must occur within the next N terms (and certainly this bound could be decreased)
However, due to the decreasing frequency of -1 in the sequence, I would be tempted to say that there is some intuitive way in which this sequence converges to 1. Is there a different notion of convergence that captures the way in which this sequence behaves?
 A: Most methods of summing divergent series can be adapted to making divergent sequences converge.  For instance, Cesaro summation replaces the $n^\text{th}$ term of a series with the mean of the first $n$ terms.  You could do the same: replace the $n^\text{th}$ term of your sequence with the mean of the first $n$ terms.  Then you would be able to show convergence to $1$ (after this replacement) in the normal sense.
Abel summation would be adapted to replace the $N^\text{th}$ term in your series with $\lim_{z \rightarrow 1^-} \sum_{n=0}^N a_n z^n$.
And so on for other methods.
Hardy's book, Divergent Series is a fun read with a lot of interesting stuff.
A: The Cesaro mean accomplishes something that is close to your intuition.
For the sequence $a_1,a_2,a_3,\cdots$, the Cesaro mean is the limit, if it exists, of the sequence $(b_n)$, where $b_n=\frac{a_1+\cdots+a_n}{n}$.
If the limit of $(a_n)$ is $a$, then the Cesaro mean of the sequence $(a_n)$ is $a$. But the Cesaro mean of the sequence $(a_n)$ may exist when the limit does not. The Cesaro mean of the sequence in your post is $1$, as desired.
A: As  André Nicolas wrote,
the Cesaro mean,
for which the $n$-th term
 is the average of the
first $n$ terms
will do what you want.
In both your cases,
for large $n$,
if $(a_n)$ is your sequence,
if
$b_n = \frac1{n}\sum_{k=1}^n a_k
$,
then
$b_n \to 1$
since the number of $-1$'s
gets arbitrarily small
compared to the number of $1$'s.
