Let's my way to do this, consider $a_n= \sum_{k=1}^n (-1)^{k-1}\frac{k}{4k-3}$, and then $$ \begin{align} \frac{1}{n}(a_1+a_2+\cdots+a_n)&=\frac{1}{n}\sum_{k=1}^{n}\sum_{j=1}^k (-1)^{j-1}\frac{j}{4j-3}\\ &=\frac{1}{n}\sum_{j=1}^n (-1)^{j-1}\frac{j}{4j-3}\sum_{k=j}^n1\\ &=\sum_{j=1}^n(-1)^{j-1}\frac{j}{4j-3}\left(1-\frac{j-1}{n}\right) \end{align} $$
I try to write the last one as the partial sum of the form of Riemann integral, but I can't find a good way to write it.
Cesàro-summability preserves additivity and extends the usual notion of convergence. Since $$ \sum_{k\geq 1}(-1)^{k-1}\frac{1}{4}\stackrel{\text{Cesàro}}{=}\frac{1}{8} \tag{1}$$ we have: $$ \sum_{k\geq 1}(-1)^{k-1}\frac{k}{4k-3}\stackrel{\text{Cesàro}}{=}\frac{1}{8}+\frac{3}{4}\sum_{k\geq 1}\frac{(-1)^{k-1}}{4k-3} \tag{2}$$ where the last series is conditionally convergent (in the usual sense) to: $$ \sum_{k\geq 1}(-1)^{k-1}\int_{0}^{1}x^{4k-2}\,dx = \int_{0}^{1}\sum_{k\geq 1}(-1)^{k-1}x^{4k-2}\,dx = \int_{0}^{1}\frac{x^2}{1+x^4}\,dx\tag{3} $$ and the claim boils down to showing that $$ \int_{0}^{1}\frac{dx}{(1+x^4)^2}=\frac{1}{32} \left(4+3 \sqrt{2} \left(\pi -\log\left(3-2 \sqrt{2}\right)\right)\right) \tag{4}$$ not that difficult through partial fraction decomposition. In a single step, $$\begin{eqnarray*} \sum_{k\geq 1}(-1)^{k-1}\frac{k}{4k-3}&\stackrel{\text{Cesàro}}{=}&\sum_{n\geq 0}(-1)^n\frac{(n+1)}{(4n+1)}\\&\stackrel{\text{Cesàro}}{=}&\int_{0}^{1}\sum_{n\geq 0}(-1)^n (n+1) x^{4n}\,dx = \int_{0}^{1}\frac{dx}{(1+x^4)^2}.\tag{5}\end{eqnarray*}$$