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$\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast\underbrace{\left[y_{\sigma1}\ast\cdots\ast\underbrace{\displaystyle \left[x,y_{\sigma i}\right] }_{\substack{=x\ast _{\displaystyle\substack{\displaystyle =x\ast y_{\sigma i}-y_{\sigma i}\astx\\\text{(since x\\\displaystyle \text{(since the inclusion of }\mathfrak{L}\\\text{into \mathfrak{L}\\\displaystyle \text{into }\operatorname*{Sym}\nolimits^{\ast}\mathfrak{L}\text{ is a morphism}\\\text{of morphism}\\\displaystyle \text{of Lie algebras)} $=\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast\underbrace{\left(y_{\sigma1}\ast\cdots\ast\underbrace{\displaystyle \left(}_{\substack{=x\ast _{\substack{\displaystyle =x\ast y_{\sigma i}\ast\cdots\ast y_{\sigma n}-y_{\sigma i}\astx\ast\cdots\ast y_{\sigma n}\\\text{(by n}\\\displaystyle \text{(by the induction hypothesis, since }i>1\text{)}}}$ $= \sum\limits_{i>0}\left( n-\left( i+1\right) +1\right) \underbrace{y_{\sigma1}\ast\cdots\ast underbrace{\displaystyle y_{\sigma1}\ast\cdots\ast x\ast y_{\sigman}}_{\displaystyle =y_{\sigma1}\ast\cdots\ast
$=\underbrace{\left( =\underbrace{\displaystyle \left( \dfrac{1}{n!}-\dfrac{1}{\left( n+1\right) !}\right)}_{=\dfrac{n}{\left( _{\displaystyle =\dfrac{n}{\left( n+1\right) !}}\sum\limits_{\sigma}x\ast y_{\sigma1}\ast        1 [made Community Wiki] Here is an explanation Pavel Etingof has given to me in email. Thanks Pavel! Every$\sigma\in S_{n}$satisfies$\sum\limits_{i}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast\left[ x,y_{\sigma i}\right] \ast\cdots\ast y_{\sigma n}=n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast\left[ x,y_{\sigma i}\right] \ast\cdots\ast y_{\sigma n}$. But since $\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast\underbrace{\left[ x,y_{\sigma i}\right] }_{\substack{=x\ast y_{\sigma i}-y_{\sigma i}\ast x\\\text{(since the inclusion of }\mathfrak{L}\\\text{into }\operatorname*{Sym} \nolimits^{\ast}\mathfrak{L}\text{ is a morphism}\\\text{of Lie algebras)} }}\ast\cdots\ast y_{\sigma n}$ $=\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast\underbrace{\left( x\ast y_{\sigma i}-y_{\sigma i}\ast x\right) \ast\cdots\ast y_{\sigma n} }_{\substack{=x\ast y_{\sigma i}\ast\cdots\ast y_{\sigma n}-y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}\\\text{(by the induction hypothesis, since }i>1\text{)}}}$$=\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast x\ast y_{\sigma i}\ast\cdots\ast y_{\sigma n}-\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$$= \sum\limits_{i>0}\left( n-\left( i+1\right) +1\right) \underbrace{y_{\sigma1}\ast\cdots\ast x\ast y_{\sigma \left(i+1\right)}\ast\cdots\ast y_{\sigma n}}_{=y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}}$$-\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$(here we substituted$i+1$for$i$in the first sum)$=\sum\limits_{i>0}\left( n-\left( i+1\right) +1\right) y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}-\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$$=\left( n-1\right) y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}-\sum\limits _{i>1}y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$, this becomes$\sum\limits_{i}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast\left[ x,y_{\sigma i}\right] \ast\cdots\ast y_{\sigma n}=n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\left( n-1\right) y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}-\sum\limits_{i>1}y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}=n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}-y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}-\sum\limits_{i>1}y_{\sigma1}\ast \cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}=n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}-\sum\limits_{i>0}y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$. Thus, (1.3.7.7) rewrites as$\dfrac{1}{n!}x\ast\sum\limits_{\sigma}y_{\sigma1}\ast\cdots\ast y_{\sigma n}=\left( \text{symmetrized product of }x,y_{1},...,y_{n}\right) +\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}-\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}\sum\limits_{i>0}y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$. Since$\left( \text{symmetrized product of }x,y_{1},...,y_{n}\right) =\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}\left( \sum\limits_{i>0}y_{\sigma1} \ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}+x\ast y_{\sigma 1}\ast\cdots\ast y_{\sigma n}\right) $, this simplifies to$\dfrac{1}{n!}x\ast\sum\limits_{\sigma}y_{\sigma1}\ast\cdots\ast y_{\sigma n} =\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}x\ast y_{\sigma1}\ast\cdots\ast y_{\sigma n}+\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$. Thus$\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}=\dfrac{1}{n!}x\ast\sum\limits_{\sigma}y_{\sigma1}\ast\cdots\ast y_{\sigma n} -\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}x\ast y_{\sigma1}\ast\cdots\ast y_{\sigma n}$$=\underbrace{\left( \dfrac{1}{n!}-\dfrac{1}{\left( n+1\right) !}\right) }_{=\dfrac{n}{\left( n+1\right) !}}\sum\limits_{\sigma}x\ast y_{\sigma1}\ast \cdots\ast y_{\sigma n}=\dfrac{n}{\left( n+1\right) !}\sum\limits_{\sigma}x\ast y_{\sigma1}\ast\cdots\ast y_{\sigma n}$. Divide this by$\dfrac{n}{\left( n+1\right) !}$to obtain$\sum\limits_{\sigma}\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\sum\limits_{\sigma}y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}=\sum\limits_{\sigma}x\ast y_{\sigma1}\ast\cdots\ast y_{\sigma n}$. In other words,$0=\sum\limits_{\sigma}\left( x\ast y_{\sigma1}\ast\cdots\ast y_{\sigma n} -y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}-\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}\right) =\sum\limits_{\sigma}\left\lbrace x,y_{\sigma1},...,y_{\sigma n}\right\rbrace =\left( n-1\right) !\sum\limits_{i}\left\lbrace x,y_{i},y_{1},...,\widehat{y_{i}},...,y_{n} \right\rbrace $(here we used that$\left\lbrace x_{1},...,x_{n+1}\right\rbrace $is symmetric in the last$n-1$variables, so that every$\sigma\in S_{n}$satisfies$\left\lbrace x,y_{\sigma1},...,y_{\sigma n}\right\rbrace =\left\lbrace x,y_{i},y_{1} ,...,\widehat{y_{i}},...,y_{n}\right\rbrace $for$i=\sigma1$). Thus,$\sum\limits_{i}\left\lbrace x,y_{i},y_{1},...,\widehat{y_{i}},...,y_{n}\right\rbrace =0\$, qed.