# The Taylor expansion of the metric at the origin in geodesic coordinates

It is well known that in geodesic coordinates we have $$g_{ij}=\delta_{ij}-\frac{1}{3}\sum_{k,l}R_{ijkl}x^{k}x^{l}+O(|x|^{3})$$ I have been trying to find a rigorous proof of it, but I cannot find a readable proof online (see this one, for example) or derive it myself. From the compatibility with metric we have $$\partial_{i}g_{jk}=\sum_{\alpha}(\Gamma^{\alpha}_{ij}g_{ak}+\Gamma^{\alpha}_{ik}g_{aj})$$ Since we know that $\nabla_{X}X=0$ for any vector from the origin, we have $\nabla_{i}\partial_{j}=0,\forall i,j$. This then implies all $\Gamma^{k}_{ij}=0$. So the first order derivative vanishes. However, it is not clear to me how to compute the second derivative. We have: $$\partial_{l}\partial_{i}g_{jk}=\sum_{\alpha}(\partial_{l}\Gamma^{\alpha}_{ij}g_{\alpha k}+\partial_{l}\Gamma^{\alpha}_{ik}g_{\alpha j})$$ This can be simplified further by noticing that in the origin we have $g_{ij}=\delta_{ij}$. Therefore the above sum is in fact $$\partial_{l}\partial_{i}g_{jk}=\partial_{l}\Gamma^{k}_{ij}+\partial_{l}\Gamma^{j}_{ik}$$ And we would have proved the statement if we can show that $$\frac{1}{2}(\partial_{l}\Gamma^{k}_{ij}+\partial_{l}\Gamma^{j}_{ik})=-\frac{1}{3}(R_{ijkl}+R_{ljki})$$ By definition we have $$R_{abcd}=g_{ae}R^{e}_{bcd}=\sum_{e}g_{ae}R^{e}_{bcd}=R^{a}_{bcd}=\partial_{c}\Gamma^{a}_{db}-\partial_{d}\Gamma^{a}_{bc}$$ Therefore we have $$R_{ijkl}+R_{ljki}=\partial_{k}\Gamma^{i}_{jl}-\partial_{l}\Gamma^{i}_{jk}+\partial_{k}\Gamma^{l}_{ij}-\partial_{i}\Gamma^{l}_{jk}$$ But I do not know why we would have $$\frac{1}{2}(\partial_{l}\Gamma^{k}_{ij}+\partial_{l}\Gamma^{j}_{ik})=-\frac{1}{3}(\partial_{k}\Gamma^{i}_{jl}-\partial_{l}\Gamma^{i}_{jk}+\partial_{k}\Gamma^{l}_{ij}-\partial_{i}\Gamma^{l}_{jk})$$ I do not know if I missed something obvious like using the Bianchi identity.