First of all the series converges for all $x$ in $[-\pi,\pi]$. Therefore it is the Fourier series of its sum function. It is continuous except at $x = 0$.
Moreover since ${a_n} = \frac{1}{{\sqrt n }}$ is not of O(1/n), (i.e. $n{a_n}$ is not bounded) the Fourier series does not converge boundedly, consequently the sum function is unbounded and hence not Riemann integrable. It is nevertheless Lebesgue integrable.
See Theorem 14 in Fourier Cosine and Sine Series.
$\newcommand{\Vector}[1]{\mathbf{#1}}\newcommand{\vece}{\Vector{e}}$The linked questions provide good answers, but may be at the technical end of "intuitive". Here's a fast-and-loose conceptual motivation:
If $\bigl(V, \langle\ ,\ \rangle\bigr)$ is an $N$-dimensional real inner product space, and if $\{\vece_{n}\}_{n=1}^{N}$ is an (ordered) orthonormal basis, then an arbitrary vector $v$ in $V$ may be written as a linear combination
$$
v = \sum_{n=1}^{N} \langle v, \vece_{n}\rangle \vece_{n}.
\tag{1}
$$
Indeed, $\{\vece_{n}\}$ is a basis of $V$, so there exist real coefficients $a_{k}$ such that
$$
v = \sum_{k=1}^{N} a_{k} \vece_{k}.
\tag{2}
$$
Taking the inner product of each side with $\vece_{n}$ gives
$\langle v, \vece_{n}\rangle = a_{n}$ because the basis $\{\vece_{n}\}$ is orthonormal.
Loosely, one might expect a similar conclusion to hold if $V$ is infinite-dimensional. Getting the definitions and hypotheses right, and proving a version of (1) in this new setting, is why any "honest" answer is bound to be technical. Phrases in quotes below are not mathematically correct, and therefore require careful inspection and/or justification.
Intuitively, let $L > 0$ be real, let $V$ be "the space of real-valued functions" on $[-L, L]$, and define an "inner product" by
$$
\langle f, g\rangle = \frac{1}{L} \int_{-L}^{L} f(t) g(t)\, dt.
$$
The functions
$$
C_{n}(t) = \begin{cases}
1/\sqrt{2}, & n = 0, \\
\cos(n\pi t/L), & n > 0;
\end{cases}\qquad
S_{n}(t) = \sin(n\pi t/L),\quad n > 0;
$$
turn out (by elementary calculus and trigonometry) to form an "orthonormal basis" of $V$.
Loosely, we expect that if $f$ is a function, we can express $f$ as an infinite sum of these basis functions, and the coefficients are the inner products of $f$ with the basis elements, i.e. (for $n > 0$),
\begin{align*}
a_{0} &= \langle f, 1\rangle = \frac{1}{L} \int_{-L}^{L} f(t)\, dt, \\
a_{n} &= \langle f, C_{n}\rangle = \frac{1}{L} \int_{-L}^{L} f(t) \cos(n\pi t/L)\, dt, \\
b_{n} &= \langle f, S_{n}\rangle = \frac{1}{L} \int_{-L}^{L} f(t) \sin(n\pi t/L)\, dt, \\
f(t) &= \frac{a_{0}}{2} + \sum_{n=1}^{\infty} (a_{n} C_{n}(t) + b_{n} S_{n}(t), \\
&= \frac{a_{0}}{2} + \sum_{n=1}^{\infty} a_{n} \cos(n\pi t/L) + b_{n} \sin(n\pi t/L).
\end{align*}
(The "special" factor of $1/2$ on the constant term arises because $C_{0} = 1/\sqrt{2} \neq 1$.)
Best Answer
The series converges pointwise to an even function $f$ on $[-\pi,\pi]\setminus\{0\}$.
By the Dirichlet test, the series is uniformly convergent on any interval $[\delta,\pi]$ where $0 < \delta < \pi$. Furthermore, we have $f$ continuous on $[\delta,\pi]$. Thus, we can integrate termwise to obtain
$$ \int_\delta^\pi f(x) \, dx = \sum_{n=2}^\infty \frac{1}{\ln n}\int_\delta^\pi \cos nx \, dx = - \sum_{n=2}^\infty \frac{\sin n\delta}{n \ln n}$$
By a well-known theorem, the series on the RHS converges uniformly on $[0,\pi]$ since the coefficients $b_n = 1/(n \ln n)$ are monotonically decreasing and satisfy $nb_n \to 0$ as $n \to \infty$.
Therefore, we can interchange the limit as $\delta \to 0$ with the sum to obtain
$$\int_0^\pi f(x) \, dx = - \sum_{n=2}^\infty \lim_{\delta \to 0}\frac{\sin n\delta}{n \ln n}=0$$
This proves that $f$ is integrable on $[0,\pi]$ as well as $[-\pi,\pi]$ since it is even.
By a similar argument we can show that
$$\frac{2}{\pi}\int_0^\pi f(x) \cos nx \, dx = \frac{1}{\ln n}$$
Therefore, this is a Fourier series for $f$.