Definite Sum of Confluent Hypergeometric involving power function

Question

I find it difficult to evaluate the following definite sum:

$$ \sum _{k=1}^K \frac{_1F_1[k,1,x]} {2^k} $$

Thank you for your time

Answer 1

In view of many other analogous questions of the OP, it is difficult to expect a satisfactory answer here. As I said in my comment, the result is a $(K-1)$th degree polynomial times $e^x$. I don't understand what can be simpler nor what can be simplified but will try to make this claim more explicit below.

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First note that $$_1F_1(k,1,x)=e^x _1F_1(1-k,1,-x)=e^{x}L_{k-1}(-x),$$ where $L_k(x)$ denotes $k$th Laguerre polynomial. So the sum of the question is \begin{align} S_K=e^{x}\sum_{k=1}^K\frac{L_{k-1}(-x)}{2^k}&=\frac12 e^x\sum_{n=0}^{K-1}\frac{L_n(-x)}{2^n}=\\ &=\frac12 e^x\sum_{n=0}^{K-1}\sum_{k=0}^n\frac{x^k}{2^{n}k!}{n \choose k}=\\ &=\frac12 e^x\sum_{k=0}^{K-1}\frac{x^{k}}{k!}\sum_{n=k}^{K-1}\frac{1}{2^n}{n \choose k}=\\ &=\frac12 e^x\sum_{k=0}^{K-1}\frac{x^{k}}{k!}\left(2-2^{1-K}{K \choose k}{}_2F_1(-k,1,K-k+1;-1)\right). \end{align} Thus we obtain a polynomial representation with explicit coefficients.

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P.S. Of course, as $K\rightarrow\infty$, a simpler representation can be obtained using the generating function of Laguerre polynomials $$G(x,t)=\frac{1}{1-t}\exp\left\{-\frac{xt}{1-t}\right\}=\sum_{n=0}^{\infty}L_n(x)t^n.$$ Namely: $$ S_{\infty}=\frac{e^{x}}{2}G\left(-x,\frac12\right)=e^{2x}.$$ However, for finite $K$ "exponential $\times$ polynomial" can only be "exponential $\times$ polynomial". It cannot eventually become a sine of $18^{\circ}$ or square root of $x^2+1$ or anything else like that.