Use epsilon-N deﬁnition of a convergent sequence to prove that the series diverges I'm stuck on this question -
Use the ε-δ deﬁnition of a convergent sequence to prove that a series is divergent. I've attached the exact question but I have no idea what to do especially since it's telling me to use the definition of a convergent sequence on a series that has no limit at infinity (it oscillates). Any help would be appreciated

 A: Let $s_n = \sum^n_{k=1}(-1)^k$.  Then, {$s_n$}$_{n\ge1}$ is the corresponding sequence.  Therefore, the series converges if and only if the sequence converges $\forall \epsilon>0$ (using the standard definition of convergent sequence).  Assume the sequence converges to some $s \in \mathbb{R}$.  Then, $\exists \delta > 0 $ such that $|s_m - s| < \epsilon$,  $\forall m > \delta$.  Let $m$ be even, then $s_m = \sum^m_{k=1}(-1)^k = 0 \Rightarrow |s_m - s| = |-s| = s < \epsilon$ so that we have $0 \le s < \epsilon$.  Since $\epsilon > 0$ is arbitrary, it follows that $s=0$.  Now consider $m+1$ which is odd, $s_{m+1} = \sum^{m+1}_{k=1}(-1)^k = 1 \Rightarrow |s_{m+1} - s| = |1 - 0| = 1 < \epsilon$ which is a contradiction for sufficiently small $\epsilon$.  Hence, the sequence {$s_n$}$_{n\ge1}$ does not converge and so $\sum^{\infty}_{k=1}(-1)^k$ does not converge.
A: The partial sums $s_n:=\sum_{k=1}^n (-1)^k$ are alternately $-1$ and $0$. So we have to prove that the sequence $(s_n)_{n\geq1}=(-1,0,-1,0,-1,0,\ldots)$ diverges.
Let an arbitrary "trial" $s\in{\mathbb R}$ be given. We shall show that $\lim_{n\to\infty} s_n=s$ cannot hold. 
When $s\leq-{1\over2}$ then  $|s_n-s|\geq{1\over2}$ for all even $n$, and this implies that the convergence condition ("there exists $n_0$ such that $\ldots$") is violated for $\epsilon={1\over3}$. Similarly, if $s\geq-{1\over2}$ we have $|s_n-s|\geq{1\over2}$ for all odd $n$, and the convergence condition is violated for $\epsilon={1\over3}$ as well.
