I already showed this method in the previous problem. You are trying to apply new knowledge to the wrong problem. Calculating such a series directly will not be easier than the integral, so why the replacement? Let's look at the problem
Let $g : \mathbb{R} \longrightarrow \mathbb{R}$ be an integrable and periodic function with period $T$ that does not change sign on the interval $[0, T]$. Let $f : [0, T] \longrightarrow \mathbb{R}$ also be a continuous function. Prove that
$$\lim_{n \to \infty } \int\limits_{0}^{T}f(x)g(nx)\, \mathrm{d}x=\frac{1}{T}\int\limits_{0}^{T}f(x)\, \mathrm{d}x\int\limits_{0}^{T}g(x)\, \mathrm{d}x$$
Proof
Let's make a change of variable $nx = t$. Then
$$\int\limits_0^T f(x)g(nx) \, \mathrm{d}x = \sum_{k=1}^n \int\limits_{\frac{(k-1)T}{n}}^{\frac{kT}{n}} f(x)g(nx) \, \mathrm{d}x = \frac{1}{n} \sum_{k=1}^n \int\limits_{(k-1)T}^{kT} f\left(\frac{t}{n}\right) g(t) \, \mathrm{d}t \tag{1}$$
Since $g$ does not change sign, for each $k$ there exists a point $c_k \in \left(\frac{(k-1)T}{n}, \frac{kT}{n}\right)$ such that
$$\int\limits_{(k-1)T}^{kT} f\left(\frac{t}{n}\right) g(t) \, \mathrm{d}t = f(c_k) \int\limits_{(k-1)T}^{kT} g(t) \, \mathrm{d}t, \tag{2}$$
which follows from the mean value theorem for integrals. In addition, due to the periodicity of $g$ with period $T$, it is satisfied
$$\int\limits_{(k-1)T}^{kT} g(t) \, \mathrm{d}t = \int\limits_0^T g(t) \, \mathrm{d}t$$
for all $k$. Substituting $(2)$ into $(1)$, we obtain
$$\int\limits_0^T f(x)g(nx) \, \mathrm{d}x = \frac{1}{n} \sum_{k=1}^n f(c_k) \int\limits_0^T g(t) \, \mathrm{d}t \tag{3}$$
Note that $\frac{T}{n} \sum_{k=1}^n f(c_k)$ is the Riemann sum of the function $f$ on the interval ([0, T]), therefore
$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n f(c_k) = \frac{1}{T} \int\limits_0^T f(t) \, \mathrm{d}t \tag{4}$$
The result itself follows from $(3)$ and $(4)$
Let's apply this result to our problem
\begin{aligned}
\lim_{n\to\infty}\int\limits_{\pi}^{2\pi}\frac{|\sin (nx) +\cos (nx)|}{ x} \ \mathrm{d}t &= \lim_{n\to\infty}\int\limits_0^{\pi} \frac{|\sin (nx) +\cos (nx)|}{ x+\pi} \ \mathrm{d}t \\
&= \frac{1}{\pi}\int\limits_0^{\pi}\frac{\mathrm{d}t}{x+\pi}\int\limits_0^{\pi}|\sin x + \cos x| \ \mathrm{d}t = \frac{2\sqrt{2}\ln 2}{\pi}
\end{aligned}
As for your attempt, here you don’t need to calculate the series, but to evaluate it!
$$\frac{1}{\pi(k+1)}\le \frac{1}{t+\pi k}\le \frac{1}{\pi k}\Rightarrow \frac{1}{\pi}\sum_{n+1}^{2n}\frac{1}{k}\le \sum_{n}^{2n-1}\frac{1}{t+\pi k}\le \frac{1}{\pi k}$$
I think this approach is more complicated than the first method. You have to use the squeeze theorem, then we get that the central sum tends to $\ln 2/\pi$ uniformly in $t$. Then all that remains is to calculate the integral
$$\int\limits_{0}^{\pi}\varphi(t)\, \mathrm{d}t=2\sqrt{2}$$
I would like to tell you for the future that it is possible to move from the limit to the integral. You can try to solve this problem yourself
True or false: If $g : \mathbb{R} \longrightarrow \mathbb{R}$ is an integrable and periodic function with period $T$, and if $\lim\limits_{n \to \infty} g(nT)$ exists, then
$$\lim_{n \to \infty} g(nT) = \frac{1}{T} \int\limits_0^T g(x) \, \mathrm{d}x\quad ? $$
