I'd like to have a closed formula for $\sum\limits_{i=0}^n e^{\sqrt{i}x}=1+e^{\sqrt{1}x}+...+e^{\sqrt{n}x}$, something similar to the formula $\sum\limits_{i=0}^ne^{ix}=\frac{e^{(n+1)x}-1}{e^x-1}$. The main problem is that the last sum is geometric, while the first one isn't. I tried some sum packages in Maple, but it seems most of the theory is geared towards hypergeometric sums, which the first sum above isn't.
Any help would be tremendously appreciated. Thanks!
There is no formula (even using special functions) but you could use tbe bounds $$I_1=\int_0^{n} e^{\sqrt{i}x}di<\sum\limits_{i=0}^n e^{\sqrt{i}x} <\int_0^{n+1} e^{\sqrt{i}x}di=I_2$$
$$I_1=\frac{2 e^{x\sqrt{n} } \left(x\sqrt{n} -1\right)+2}{x^2}$$ $$I_2=\frac{2 e^{x\sqrt{n+1} } \left(x\sqrt{n+1} -1\right)+2}{x^2}$$
Trying for $x=\frac 12$ and a few values of $n$ $$\left( \begin{array}{ccccc} n & I_1 & I_2 & \frac 12(I_1+I_2) & \text{summation} \\ 5 & 10.8884 & 14.1190 & 12.5037 & 12.8314 \\ 10 & 30.5969 & 35.6515 & 33.1242 & 33.4443 \\ 15 & 59.9519 & 67.1124 & 63.5322 & 63.8415 \\ 20 & 100.522 & 110.143 & 105.332 & 105.629 \\ 25 & 154.190 & 166.680 & 160.435 & 160.717 \\ 30 & 223.108 & 238.930 & 231.019 & 231.285 \\ 35 & 309.698 & 329.369 & 319.533 & 319.781 \\ 40 & 416.659 & 440.755 & 428.707 & 428.934 \\ 45 & 546.993 & 576.151 & 561.572 & 561.777 \\ 50 & 704.021 & 738.946 & 721.483 & 721.664 \\ 55 & 891.409 & 932.878 & 912.143 & 912.296 \\ 60 & 1113.19 & 1162.06 & 1137.63 & 1137.75 \\ 65 & 1373.81 & 1431.02 & 1402.42 & 1402.51 \\ 70 & 1678.14 & 1744.70 & 1711.42 & 1711.47 \\ 75 & 2031.49 & 2108.55 & 2070.02 & 2070.03 \\ 80 & 2439.70 & 2528.48 & 2484.09 & 2484.06 \\ 85 & 2909.14 & 3010.97 & 2960.05 & 2959.98 \\ 90 & 3446.73 & 3563.08 & 3504.90 & 3504.78 \\ 95 & 4060.02 & 4192.47 & 4126.25 & 4126.07 \\ 100 & 4757.22 & 4907.50 & 4832.36 & 4832.12 \end{array} \right)$$
So, it seems that
$$\frac 12(I_1+I_2)=\frac{e^{x\sqrt{n} } \left(x\sqrt{n} -1\right)+e^{x\sqrt{n+1} } \left(x\sqrt{n+1}-1\right)+2}{x^2}$$ could be a decent approximation.