I need your help about a series : $$ \forall n \in \mathbb{N}^* \backslash\left\{ 1 \right\} \mbox{, } \sum_{\sigma \in S_n} \sigma(1) $$ My goal is to give an explicit formula for this series. I tried to understand this with $n = 2$ and $n = 3$. For $n = 2$ we have identity and transposition in $S_2$ so
$$ \sum_{\sigma \in S_2} \sigma(1) = \mathrm{id}(1) + \tau(1) = 1 + 2 = 3 $$
For $n = 3$, we have identity, 3 transposition and 2 $3$-cycle in $S_3$ so
$$ \sum_{\sigma \in S_3} \sigma(1) = \mathrm{id}(1) + \tau_1(1) + \tau_2(1) + \tau_3(1) + c_1(1) + c_2(1) = 1 + (1+2+3) + (2+3) = 2(1+2+3)=12 $$
I thought this series links to $\sum_{k = 1}^n k$ because it works for $n = 2$ but fails for $n = 3$... I'm here to ask you an indication, please. Thank you.
For every $i = 1, ..., n$, let $T_i = \lbrace \sigma \in S_n \mid \sigma(1)=i \rbrace$.
One has $$\sum_{\sigma \in S_n} \sigma(1) = \sum_{i=1}^n \sum_{\sigma \in T_i} \sigma(1) = \sum_{i=1}^n \sum_{\sigma \in T_i} i = \sum_{i=1}^n i \times \mathrm{Card}(T_i)$$
Moreover, for every $i=1, ..., n$, one has $\mathrm{Card}(T_i)=(n-1)!$. Indeed, the following map is bijective : $$\varphi : \left|\begin{array}{ccc} T_i &\rightarrow & S_{n-1} \\ \sigma & \mapsto &\tau\end{array}\right.$$
where $\tau$ is defined by $\tau(k) = \left\lbrace\begin{array}{cl} 1 & \text{ if } k=\sigma^{-1}(1)-1 \\ \sigma(k+1)-1 & \text{ otherwise}\end{array}\right.$
So one deduces that $$\sum_{\sigma \in S_n} \sigma(1) = (n-1)! \sum_{i=1}^n i = (n-1)! \dfrac{n(n+1)}{2}$$
i.e. $$\boxed{\sum_{\sigma \in S_n} \sigma(1) = \dfrac{(n+1)!}{2}}$$