The third covariant of a tensor in Riemannian manifolds I'm reading Petersen's book "Riemannian Geometry 3rd", No.171 in GTM. On page 82, there's a formula reads:
$$\nabla^3_{X,Y,Z}W=\nabla^2_{X,Y}(\nabla_ZW)-\nabla_{\nabla^2_{XY}Z}W$$
I'm quite puzzled with this formula. On the right hand site, it doesn't seem to be tensorial in the "Z" position. Is this formula false?
 A: Yeah I don't think that formula works as written; there are some missing terms.
The good news is that the omitted terms don't matter in context. The purpose of the formula in the context of the book was to compute the difference $\nabla^3_{X,Y,Z} W - \nabla^3_{Y,X,Z} W$. The omitted terms are symmetric in $X$ and $Y$, so they cancel out.

The rest of this answer is just the details. My claim is that:
$$
  \nabla^2_{X,Y}(\nabla_Z W)
  = \nabla^3_{X,Y,Z} W
    + \nabla_{\nabla^2_{X,Y} Z} W
    + \nabla^2_{X,\nabla_Y Z} W
    + \nabla^2_{Y,\nabla_X Z} W
$$
Note the last two terms, which didn't appear in the original formula.
I'll do the computation in index notation, but I'm sure you can do it at home in whatever notation you prefer. We can translate $\nabla^2_{X,Y}(\nabla_Z W)$ to index notation and expand using the product rule twice:
\begin{align}
  & X^i Y^j \nabla_i \nabla_j (Z^k \nabla_k W^\ell) \\
  &= X^i Y^j \nabla_i (
    \nabla_j Z^k \nabla_k W^\ell
    + Z^k \nabla_j \nabla_k W^\ell
    ) \\
  &= X^i Y^j (
    \nabla_i \nabla_j Z^k \nabla_k W^\ell
    + \nabla_j Z^k \nabla_i \nabla_k W^\ell
    + \nabla_i Z^k \nabla_j \nabla_k W^\ell  
    + Z^k \nabla_i \nabla_j \nabla_k W^\ell
    )
  \end{align}
In this, the first term is $\nabla_{\nabla^2_{X,Y} Z} W$, the second term is $\nabla^2_{X,\nabla_Y Z} W$, the third term is $\nabla^2_{Y,\nabla_X Z} W$, and the last term is $\nabla^3_{X,Y,Z} W$.
