Do the Christoffel symbols $\Gamma_{rn}^w\partial_sV_w = \Gamma_{sn}^w\partial_rV_w$? In lecture 3 (about 97 min into the lecture) of Leonard Susskind's general relativity course, he suggests finding the Riemann curvature tensor in terms of the Christoffel symbols as an exercise.
I started from $$D_r D_s V_n - D_s D_r V_n = R_{rs}^t{_n} V_t$$ as a definition of the Riemann curvature tensor. 
The covariant derivative is defined by $$D_sV_n = \partial_sV_n - \Gamma_s^u{_n}V_u$$ and the Christoffel symbols are defined in terms of the metric by $$\Gamma_s^u{_n} = \dfrac12 g^{ut}\left( \partial_sg_{tn} + \partial_n g_{st} - \partial_tg_{sn}\right).$$
I'm supposed to find $$R_{rs}^t{_n} = -\partial_r\Gamma_s^t{_n} + \partial_s\Gamma_r^t{_n} + \Gamma_r^w{_n}\Gamma_s^t{_w} - \Gamma_s^w{_n}\Gamma_r^t{_w}.$$
I get that this works, but only if $$\Gamma_r^w{_n}\partial_sV_w = \Gamma_s^w{_n}\partial_rV_w.$$
I'm stuck showing this. I tried expanding the Christoffel symbols out into the metric, but I wind up trying to show things like $$\partial_r g_{tn}\partial_s = \partial_s g_{tn}\partial_r,$$ which I don't see how to justify.
(It may also be that I'm making algebra errors because I'm unfamiliar with tensor calculus, so any advice you have would be helpful here.)
 A: I'm over a year late to answer this question, but here goes. I think you must have misapplied some identities or got lost in the weeds during your derivation. This has to do with the torsion-freeness of the connection. We have, using the formulas for the covariant derivative of a rank 2 tensor:
\begin{align*}
D_rD_sV_n=&D_r(\partial_s V_n-\Gamma^u_{sn}V_u)\\
&=\partial_r \partial_s V_n-(\partial_r \Gamma^u_{sn})V_u-\Gamma^u_{sn}\partial_r V_u\\
&-\Gamma^a_{rs}(\partial_a V_n-\Gamma^u_{an}V_u)\\
&-\Gamma^a_{rn}(\partial_s V_a-\Gamma^u_{sa}V_u)
\end{align*}
We want to antisymmetrize with respect to r and s, and I want to use bracket notation $V_{[a} U_{b]}\equiv \frac{1}{2}(V_a U_b-V_b U_a)$ for brevity. To do this without ambiguity, we should try to rearrange all our indices to put r and s next to each other. Using the symmetry of the Christoffel symbol in the lower two indices (a consequence of torsion freeness)...
\begin{align*}
D_rD_sV_n=&-(\partial_r \Gamma^u_{sn})V_u-\Gamma^u_{ns}\partial_r V_u\\
&-\Gamma^a_{nr}\partial_s V_a+\Gamma^a_{nr}\Gamma^u_{sa}V_u\\
&+\text{symmetric stuff}
\end{align*}
Then we have:
$$
D_{[r}D_{s]}V_n=-(\partial_{[r} \Gamma^u_{s]n})V_u-\Gamma^u_{n[s}\partial_{r]} V_u-\Gamma^a_{n[r}\partial_{s]} V_a+\Gamma^a_{n[r}\Gamma^u_{s]a}V_u$$
The first and last terms give the Riemann tensor up to factors of two. The middle two terms add up to zero because by definition, $\Gamma^u_{n[s}\partial_{r]}=-\Gamma^u_{n[r}\partial_{s]}$.
What did you do  wrong?
You might have forgot to properly distribute the partial derivative across $\partial_r(\Gamma^u_{sn}V_u)$. If you work with the convention that $\partial_r$ acts on everything to the right of it, you have to be very careful about it. If you find yourself making mistakes or overusing parentheses, you can use the symbol $\overrightarrow\partial$ to mean "acts on everything to the right" and the symbol $\partial$ to mean "acts on the thing immediately to the right". So as an operator equation $\overrightarrow\partial_x f=\partial_xf+f\overrightarrow\partial_x$.
You also might have misapplied the covariant derivative of a 2-tensor. We have:
$$D_a (V_{bc})=\partial_a V_{bc}-\Gamma^u_{ab}V_{uc}-\Gamma^u_{ac}V_{bu}$$
There are also plenty of other bookkeeping / index mistakes that could be made here.
Other References
pg. 82 / equation 3.66 of Sean Carroll's lecture notes covers this, making no assumptions about torsion.
