$$\color{red}{\sum_{i=1}^{n} \frac {x^{2i-1}}{\sqrt{2i}}}$$
Finding the partial sum
$$\begin{align*}
\color{green}{\sum_{k=1}^m \frac {z^k}{k^n}}
&= \left(\sum_{k\ge1} - \sum_{k\ge m+1}\right)\frac {x^k}{k^n}
\\&=\operatorname{Li}_n(x) - \sum_{k=m+1}^\infty\frac {x^k}{k^n}
\\&=\operatorname{Li}_n(x) - \sum_{j \ge0}\frac {x^{m+1+j}}{(m+1+j)^n}
\\&=\color{green}{\operatorname{Li}_n(x) - x^{m+1}\Phi(x, n, m+1)}
\end{align*}$$
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 $k \implies b_{k+1} -b_k = 0$ or simply $b_\delta = \frac 1x$ where $\delta$ 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
$$\begin{align*}
\sum_{k=1}^n\frac {x^{2k-1}}{\sqrt{2i}}
&=S_nb_n\\
&=\frac 1x \times \sum_{k = 1}^n \frac {x^{2k}}{\sqrt{2k}}\\
&=\frac 1x \frac 1{\sqrt2}\sum_{k = 1}^n \frac {(x^2)^k}{k^\frac12}\\
&=\color{red}{\frac 1{\sqrt 2 x} \left(\operatorname{Li}_{\frac 12}(x^2) - x^{2n + 2}\Phi(x^2, \frac 12, n + 1)\right)}
\end{align*}$$