Compute (if it exists) the one-sided limit of $$\underset{x \rightarrow 1-}{\lim} \sum_{k=0}^{\infty} (-1)^k \ k \ x^k $$
I'm finding the question really confusing, especially the one-sided limit of $x$ combined with the series. I've attempted to solve the question the following way using Cesaro summability, but my answer is wrong (the limit exists). Is my choice of method wrong, or have I made some other mistake?
Let $x=1$ so we are looking for $\sum_{k=0}^{\infty} (-1)^k \ k = 0-1 +2-3+4-5+ \dots $The first few partial sums are $0,-1,1,-2,2,-3,3,-4, \dots$ which leads me to think that I should check for Cesaro summability.
Denote $s_n, \ n \geq 0$ as the partial sum. Define $$\sigma_n = \frac{s_0 + s_1 + \dots + s_n}{n+1}$$ Notice that if $n$ is even, then $\sigma_n = 0$. If $n$ is odd, then $\sigma_n = \frac{s_n}{n+1}$. Evaluating the first few $\sigma_n$ gives us $0,-\frac{1}{2}, 0, -\frac{1}{2}, 0,-\frac{1}{2}, 0, -\frac{1}{2}, \dots$. Since $\underset{n \rightarrow \infty}{\lim} \sigma_n$ does not exist, we should check if the series is $(C,2)$ summable.
Define $$\tau_n = \frac{\sigma_0 + \dots + \sigma_n}{n+1}$$ The first few $\tau$:s are $0, -\frac{1}{2}\cdot \frac{1}{2}, -\frac{1}{2}\cdot\frac{1}{3}, -\frac{1}{2}\cdot\frac{1}{4}, \dots$
So $$\tau_n = 0 + \sum_{k=1}^n \left(\frac{1}{1+k}\right)\left( -\frac{1}{2}\right) = 0 + \sum_{k=1}^n -\frac{1}{2+2k}$$ Taking the limit of $\tau$ as $n \rightarrow \infty$ shows that $\tau_n$ diverges, thus $\underset{x \rightarrow 1-}{\lim} \sum_{k=0}^{\infty} (-1)^k \ k \ x^k $ does not exist.
Note: The correct answer is $-\frac{1}{4}$
