>>40,43
In some cases, yes, the series will remain bounded despite the fact that it is always increasing.
Consider a squirrel who is standing 10 feet from a pole. Every ten minutes, the squirrel moves halfway to the pole. The squirrel's distance from it's starting point is always increasing, yet it will never move past the pole.
For a more algebraic interpretation, consider the proof that s(n) = \sum_{i=0}^{n}(1/(2^{i})) is bounded.
Note that if r > 0, then you can do the following:
1 + r + r^2 + ... + r^{n-1} + r^n
= ((1 - r)/(1 - r)) (1 + r + r^2 + ... + r^{n-1} + r^n)
= (1/(1 - r)) ((1 - r)1 + (1 - r)r + (1 - r)r^2 + ... + (1 - r)r^{n-1} + (1 - r)r^n)
= (1/(1 - r)) ((1 - r) + (r - r^2) + (r^2 - r^3) + ... + (r^{n-1} - r^{n}) + (r^n - r^{n+1}))
= (1/(1 - r)) ((1 + -r) + (r + -r^2) + (r^2 + -r^3) + ... + (r^{n-1} + -r^n) + (r^n + -r^{n+1}))
= (1/(1 - r)) (1 + (-r + r) + (-r^2 + r^2) + (-r^3 + r^3) + ... + (-r^{n-1} + r^{n-1}) + (-r^n + r^n) + -r^{n+1})
= (1/(1 - r)) (1 + 0 + 0 + 0 + ... + 0 + 0 + -r^{n+1})
= (1/(1 - r)) (1 - r^{n+1})
= (1 - r^{n+1})/(1 - r)
So \sum_{i=0}^{n}(r^i) = (1 - r^{n+1})/(1 - r)
In the special case of r = (1/2), we have:
\sum_{i=0}^{n}((1/2)^i)
= (1 - (1/2)^{n+1})/(1 - (1/2))
= (1 - (1/2)^{n+1})/(1/2)
= 2(1 - (1/2)^{n+1})
= 2 - 2(1/2)^{n+1}
= 2 - 2/(2^{n+1})
= 2 - 1/(2^{n})
So we have a closed form expression:
s(n) = \sum_{i=0}^{n}(1/(2^i))
= 2 - 1/(2^n)
The the squirrel will never pass the pole, this series will never rise above 2. And we can prove it without talking about infinity:
0 <= n
1 <= 2^n
0 < 1/(2^n) <= 1
-1/(2^n) < 0
s(n) = 2 - 1/(2^n) < 2
So c = 2 works.