Convergence of the Average of Partial Sums

Question

The Maclaurin series of $\frac{1}{x+1}$ is of course $\sum_{n=0}^{\infty}\left(-1\right)^{n}\left(x\right)^{n}$, and its interval of convergence is only considered to be $-1<x<1$. Though, it seems only 'somewhat unconverge' at $x=1$, since $\sum_{n=0}^{\infty}\left(-1\right)^{n}$ neither diverges to infinity or converges to a fixed value. One reasonable way in which it seems you could extend the interval to include that point is taking the average of the partial sum. For $x=1$ of $1/x+1$: $$\lim_{r\to\infty}\frac{\left(\sum_{n=1}^{r}\left(\sum_{i=0}^{n}\left(-1\right)^{i}\right)\right)}{r}$$ This converges to $\frac12$, which is exactly what $1/1+x$ is equal to at $x=1$. So for my first question, is this a valid way to extend the function, and, if the limit of the average of partial sums converges, will it converge to the function value of the taylor series which it represents?

Secondly, is it valid to take an average of average of partial sums, an average of average of average, etc.? For example, to find the convergence of the series of $\sum_{n=0}^{\infty}\left(-1\right)^{n}n$ you would have to take the average of the averages, or $$\lim_{w\to\infty}\frac{\left(\sum_{k=1}^{w}\left(\frac{\left(\sum_{l=1}^{k}\left(\sum_{n\ =0}^{l}\left(-1\right)^{n}n\right)\right)}{k}\right)\right)}{w}$$ The answer from this calculation gives the sum as $\frac14$. Is this a valid answer for the sum?