How to prove that the sequence $(-1)^{n+1} \times \frac {n}{n+1} $ diverges?
Solution 1:
Remember what it means for a sequence to converge: $(\forall \epsilon>0) (\exists N\in\mathbb N)( \forall n\geq N) |a_n-L|<\epsilon$.
The logical negation is $(\exists \epsilon>0) (\forall N\in\mathbb N)( \exists n\geq N) |a_n-L|\geq \epsilon$.
That is, to show a sequence does not converge to a proposed limit, it suffices to find some $\epsilon$, and some $n$ (no matter how large a lower bound for $n$ you choose) s.t. the sequence is not within $\epsilon$ of the proposed limit. Showing this for any proposed limit shows the sequence is divergent.
Thus, the proof chooses particular $\epsilon$ and particular $n$ to work for each case considered. Choosing these requires some intuition or calculations. We know the sequence starts out at $1/2$ ($n=1$) and alternates signs between positive (odd $n$) and negative (even $n$) whilst increasing in absolute value, oscillating between close to $+1$ and $-1$ in the long run. Thus,
- To show it does not converge to $0$, it suffices to choose $\epsilon=1/2$ since we know it is never within $1/2$ away from $0$.
- To show it does not converge to a positive limit, it suffices to choose $\epsilon$ equal to half the limit size since it would have to be positive to be within $\epsilon$ of the limit, but we know this is not true at even $n$.
- To show it does not converge to a negative limit, it suffices to choose $\epsilon$ equal to half the limit size since it would have to be negative to be within $\epsilon$ of the limit, but we know this is not true at odd $n$.