Let $\rho \in (0,\infty)$ and for any integer $m \ge 1$, define $\mu_m \ge 0$ by $$ \mu_m := \sum_{k=0}^{m-1}\frac{\rho^k}{k+1}{m \choose k}{m-1\choose k}. $$
Finally define $E \ge 0$ by $$ E := \sum_{m=0}^\infty \frac{\mu_m}{m!} $$
Question. Is there an analytic formula $E$ in terms of $\rho$ ?
Context: $E$ corresponds to the expected value of the trace of the exponent of an $n \times d$ Wishart matrix, in the limit $n,d \to \infty$ with $n/d \to \rho$.
$$\mu_m(\rho) := \sum_{k=0}^{m-1}\dfrac{\rho^k}{k+1}{m \choose k}{m-1\choose k}=\, _2F_1(1-m,-m;2;\rho)$$ where appears the Gaussian hypergeometric function. $$E_n(\rho)=\sum _{m=0}^{n } \frac{\, _2F_1(1-m,-m;2;\rho )}{m!}=P_{n-1}(\rho)$$
I do not think that I could go much beyond except writing
$$E_\infty(\rho)=e \sum_{n=0}^\infty \frac{a_n} {n!}\, \rho^n$$
where the first coefficients form the sequence $$\left\{1,\frac{1}{2},\frac{1}{2},\frac{13}{24},\frac{73}{120},\frac{167}{240},\frac{4051 }{5040},\frac{37633}{40320},\frac{43817}{40320},\frac{3582133632}{2827955323},\frac{5 25382200}{355805701}\right\}$$
For $\rho=\frac 1 {10}$ this gives $$\frac{4919476309996273890952908198463621}{4673674354662005734121472000000000}e=\color{red}{2.8612440756935690}0924$$ while the infinite summation gives $\color{red}{2.8612440756935690104}$.