Is it possible to simplify $$ S(\mathbf{x})=\sum_{k=0}^{n} \frac{e^{\sum_{i=0}^kx_{i}}}{\sum_{i=0}^k x_{i}} $$ A few observations:
- $\sum_{k=0}^{n}\sum_{i=0}^k x_{i}=\sum_{k=0}^{n}(n+1-k)x_k$
- $ e^{\sum_{i=0}^kx_{i}}=\prod_{i=0}^k e^{x_i}$
- $S(\mathbf{x})=\sum_{k=0}^{n}e^{\sum_{i=0}^kx_{i}-\log\left( \sum_{i=0}^k x_{i}\right)}$
Perhaps one could approximate $S(\mathbf{x})$ by defining $y\equiv \sum_{i=0}^k x_{i}$ and finding an approximation to $y-\log(y)$. Any ideas?
Edit: If $y_k=\sum_{i=0}^k x_i$ is a bounded sequence, can we hope to find an approximation of $$ \lim_{n\to\infty}\sum_{k=0}^n \frac{e^{y_k}}{y_k} $$ for $x_i>0$?
Too long for a comment
As noted by @SangchulLee : If $(x_i)$ is a constant sequence with value $a≠0$ , then the sum is
$$ S(a,n) = \frac{e^a}{a}+\frac{e^{2a}}{2a}+\cdots+\frac{e^{na}}{na}+\frac{e^{(n+1)a}}{(n+1)a} \tag{1} $$
Suppose $a >0$, then
$$ S(a,n) = \sum_{k = 1}^{n+1} \int_{-\infty}^1 e^{y k a} dy\\ = \int_{-\infty}^1 \sum_{k = 1}^{n+1} ( e^{y a})^k dy\\ = \frac{1}{a} \int_{-\infty}^a \sum_{k = 1}^{n+1} ( e^{z})^k dz\\ = \frac{1}{a} \int_{-\infty}^a \frac{e^{z} (e^{z (n + 1)} - 1)}{e^{z} - 1} dz\\ $$
Analytic solutions for this integral for fixed $n$ bring back the sum as in (1). However, the integral may be useful for finding approximations. To do so, let us factor out an even term:
$$ S(a,n) = \frac{1}{a}\int_{-\infty}^a \frac{e^{z} (e^{z (n + 1)} - 1)}{e^{z} - 1} dz\\ = \frac{1}{a}\int_{-\infty}^a \frac{e^{z (n+2)/2} \sinh (z (n+1)/2) }{\sinh (z/2)} dz\\ = \frac{2}{a} \int_{-\infty}^{a/2} \frac{e^{w (n+2)} \sinh (w (n+1)) }{\sinh (w)} dw $$
Note that $\frac{\sinh (w (n+1)) }{\sinh (w)} $ is an even function of $w$ which may be used for further approximations.