$$\sum_{i=1}^{n} \frac {x^{2i-1}}{\sqrt{2i}}$$
It is very clear for me that it has to be polylogarithm function but as it is partial sum I tried to split the series as
$$\sum_{i=1}^{\infty} \frac {x^{2i-1}}{\sqrt{2i}}- \sum_{i=n+1}^\infty\frac {x^{2i-1}}{\sqrt{2i}}$$ then $\frac {x^{2i}}{\sqrt{2i}}\frac 1x$
as $$Li_n(z) = \sum_{k=1}^{\infty}\frac{z^k}{k^n}$$
but no luck(:
@as per comment @Daarshan P.
From WolframAlpha but how should one proceed if its sum of products $\frac {x^{2i}}{\sqrt{2i}} \& \frac 1x$
$$\color{red}{\sum_{i=1}^{n} \frac {x^{2i-1}}{\sqrt{2i}}}$$
Finding the partial sum $$\begin{align*} \color{green}{\sum_{k=1}^m \frac {z^k}{k^n}} &= \left(\sum_{k\ge1} - \sum_{k\ge m+1}\right)\frac {x^k}{k^n} \\&=Li_n(x) - \sum_{k=m+1}^\infty\frac {x^k}{k^n} \\&=Li_n(x) - \sum_{j \ge0}\frac {x^{m+1+j}}{(m+1+j)^n} \\&=\color{green}{Li_n(x) - x^{m+1}\phi(x, n, m+1)} \end{align*}$$
Consider, $$\color{blue}{\sum_{k=1}^xa_kb_k \equiv S_xb_x - \sum_{k=1}^{x - 1}S_x(b_{k+1}-b_k)} \text{ Where } \color{green}{S_u = \sum_{k=1}^ua_k}$$
The blue identity is more of intuitional if we split the consecutive terms of b
here, $a_k = \frac {x^{2k}}{\sqrt{2k}} \text{ & } b_k = \frac 1x$
as $\frac 1x $ is independent of $k \implies b_{k+1} -b_k = 0 \text{ or simply } b_\delta = \frac 1x \text{ where } \delta \text{ is anything}$
hence,
$$\begin{align}\sum_{k=1}^xa_kb_k \equiv S_xb_x \text{ Where } \color{green}{S_u = \sum_{k=1}^u \frac {x^{2k}}{\sqrt{2k}}} \end{align}$$
Here, we get partial sum $$\begin{align*} \sum_{k=1}^n\frac {x^{2k-1}}{\sqrt{2i}} &=S_nb_n\\ &=\frac 1x \times \sum_{k = 1}^n \frac {x^{2k}}{\sqrt{2k}}\\ &=\frac 1x \frac 1{\sqrt2}\sum_{k = 1}^n \frac {(x^2)^k}{k^\frac12}\\ &=\color{red}{\frac 1{\sqrt 2 x} \left(Li_{\frac 12}(x^2) - x^{2n + 2}\phi(x^2, \frac 12, n + 1)\right)} \end{align*}$$