# Maximize $f(x) = \sum_{i=1}^{\infty} \frac{e^{-x}x^i}{i!}\left(\alpha+\frac{1}{\sqrt{i}}\right)(x_0-x)$

I would like to maximize the following function: \begin{align} f(x) = \sum_{i=1}^{\infty} \frac{e^{-x}x^i}{i!}\left(\alpha+\frac{1}{\sqrt{i}}\right)(x_0-x), \end{align} with $$\alpha\geq0$$ and $$0\leq x\leq x_0$$.

Differentiating I get \begin{align} \alpha(e^{-x}(x_0-x+1)-1) + \sum_{i=1}^{\infty}\frac{e^{-x}x^{i-1}}{(i-1)!\sqrt{i}}\left(\left((x_0-x\left(1-\frac{1}{i}\right)\right)-1\right). \end{align} From there I don't now where to go. Note that I would still (though less) be happy with only a lower bound on $$f(x)$$ as long as it is of the form $$\alpha x_0 - g(x_0)$$ with $$g(x_0) = o(x_0)$$.

A closed form for the minimum escapes me; however, it might be useful to notice that \begin{aligned} \alpha(1-e^{-x})(x_0-x)\leq f(x)\leq (1+\alpha)(1-e^{-x})(x_0-x) \end{aligned} $$0\leq x\leq x_0$$. The maximum of the end functions is attain at a common point $$0.