Confusion regarding delta definition of limit

[f24u7]

I don't quite get the significance of the delta limit definition,

if n>N and |s_n−s|< ϵ , why does the limit converges

does this simply means that there exist a number ε such that if n is great enough it will be greater than s by ε?

But this doesn't make sense, because s is the value the sequence converges to, so s_ncannot be greater than s

what confuses me even more is that (-1)ⁿ is a diverging sequence but
it satisfies |s_n-s| < ε for ε>1

Can someone clarify this concept, thank you

 

[jbunniii]

I don't quite get the significance of the delta limit definition,

if n>N and |s_n−s|< ϵ , why does the limit converges

does this simply means that there exist a number ε such that if n is great enough it will be greater than s by ε?

No, first of all you need to state the full definition: for EVERY ##\epsilon > 0## (no matter how small), there exists some ##N## such that ##|s_n - s| < \epsilon## for every ##n > N##. In general, choosing a smaller ##\epsilon## means that the required ##N## will be larger.

Note that ##|s_n - s| < \epsilon## means exactly that ##-\epsilon < s_n - s < \epsilon## or equivalently ##s - \epsilon < s_n < s + \epsilon##. So convergence means that no matter how small an ##\epsilon## I choose, it's possible to find an ##N## such that ##s_n## will be contained within the interval ##(s-\epsilon, s+\epsilon)## for all ##n > N##.

what confuses me even more is that (-1)ⁿ is a diverging sequence but
it satisfies |s_n-s| < ε for ε>1

You didn't mention what ##s## is in this case. But note that it's not enough for the inequality to be true for ##\epsilon > 1##. It has to be possible to make it true for any ##\epsilon > 0##, no matter how small. To see that this is impossible, take ##\epsilon = 1/2## for example, and show that there can't be any ##s## that works.