$\sum_{k=1}^{\infty} (-1)^k k$ is $(C,2)$ summable to $-\frac{1}{4}$

I am trying to show that $$\sum_{k=1}^{\infty} (-1)^k k = -\frac{1}{4}\quad (C,2)$$First I look at the partial sums:

$S_1 = -1$, $S_2 = 1$, $S_3 = -2$, $S_4 = 2$...

These will diverge to $\pm\infty$ so the series is not $(C,1)$. Now consider $\sigma_i$

$\sigma_1 = -1$, $\sigma_2 = 0$, $\sigma_3 = -\frac{2}{3}$, $\sigma_4 = 0$...

We notice that the even terms are zero. Now consider $\tau_n$ to be the sequence converging to the $(C,2)$ limit.

$\tau_1 = -1$, $\tau_2 = -\frac{1}{2}$, $\tau_3 = -\frac{-5}{9}$,$\tau_4 = -\frac{-5}{12}$...

We can notice that by removing the odd terms we get rid of the occasionally decreasing terms, which occur due to the zeroes from $\sigma$'s. Now we have an increasing "subsequence" which is bounded by $0$ and thus must converge. But I am unsure how to show the exact limit.. Any hints?:) thanks!