$$\sum_{k=1}^{\infty}{\frac{1}{\sqrt{kx}} \cdot \ln{\frac{2x + k}{x + k}}}$$
for convergence on sets $E1 = (0, 1)$ and $E2 = (1, +\infty)$.
- Explore the pointwise convergence, fix $$x_0$$
By D'Alembert, consider the ratio of members $$a_{n+1} / a_n$$,it equals to $$-\sqrt{\frac{k}{k + 1}} \cdot ln{\frac{x_0}{(x_0 + k) \cdot (x_0 + k + 1)}}$$
the first multiplier is less than one, the second is also => the product is less than 1, and the number series converges on the Dalamber basis, so the original func series converges
- prove the absence of uniform convergence on the set $E2 = (1, +\infty)$. Take the sequence $$x_n = k, m = N, n = 2N$$ to negate the Cauchy condition. We got $$\sum_{k=N + 1}^{2N}{\frac{1}{k} \cdot \ln{\frac{3}{2}}} \ge \sum_{k=N+1}^{2N}{\frac{1}{2N} \cdot \ln{\frac{3}{2}}} \ge \frac{1}{2N} \cdot ln{\frac{3}{2}} \cdot \sum_{k = N + 1}^{2N} 1 = ln{\frac{3}{2}}\cdot\frac{1}{2} = epsilon$$
selected eps for which the Cauchy criterion is incorrect => the series does not converge uniformly.
- What can we do to prove that given series converges/diverges on set E1 and am I correct in my solution about set E2?
In your argument for part (2), $x_N$ cannot depend on the summation index $k$ in the negation of the uniform Cauchy condition.
To prove non-uniform convergence of the series on $(1,\infty)$, we must show that the uniform Cauchy condition is violated in that there exists $\epsilon_0 > 0$ such that for any $N \in \mathbb{N}$ there are positive integers $m > n \geqslant N$ and $x_N \in (1,\infty)$ (note dependence here only on $N$) for which
$$|S_m(x_N)-S_n(x_N)| = \sum_{k=n+1}^m\frac{1}{\sqrt{kx_N}} \cdot \ln{\frac{2x_N + k}{x_N + k}} \geqslant \epsilon_0$$
Taking $n = N$, $m = 2N$ and $X_N = N$, we have
$$|S_{2N}(x_N) - S_N(x_N)| = \sum_{k=N+1}^{2N}\frac{1}{\sqrt{kN}} \cdot \ln{\frac{2N + k}{N + k}} \geqslant N \cdot \frac{1}{\sqrt{2N\cdot N}}\cdot \min_{N+1 \leqslant k \leqslant 2N}\ln \frac{2+ \frac{k}{N}}{1 + \frac{k}{N}}$$
Since $x \mapsto \ln \frac{2+x}{1+x}$ is decreasing for all $x >1$ it follows that
$$\min_{N+1 \leqslant k \leqslant 2N}\ln \frac{2+ \frac{k}{N}}{1 + \frac{k}{N}}= \ln \frac{2 + \frac{2N}{N}}{1 + \frac{2N}{N}}=\ln \frac{4}{3},$$
Thus,
$$|S_{2N}(x_N) - S_N(x_N)| \geqslant \frac{1}{\sqrt{2}} \ln \frac{4}{3}:= \epsilon_0$$
and the uniform Cauchy condition is violated.
Pointwise Convergence and Uniform Convergence on $(0,1)$
Note that for all $x > 0$
$$\tag{1}0\leqslant \frac{1}{\sqrt{kx}} \ln{\frac{2x + k}{x + k}}=\frac{1}{\sqrt{kx}} \ln{\frac{1+ \frac{2x}{k}}{1 + \frac{x}{k}}}=\frac{1}{\sqrt{kx}} \left[\ln\left(1 + \frac{2x}{k}\right) - \ln\left(1 + \frac{x}{k}\right)\right]$$
Applying the well-known inequality $\frac{y}{1+y} \leqslant \ln(1+y) \leqslant y$ we have
$$\ln\left(1 + \frac{2x}{k}\right) \leqslant \frac{2x}{k}, \quad \ln\left(1 + \frac{x}{k}\right)\geqslant \frac{\frac{x}{k}}{1 + \frac{x}{k}}= \frac{x}{k+x},$$
and using this to bound terms on the RHS of (1) we get
$$\tag{2}0\leqslant \frac{1}{\sqrt{kx}} \ln{\frac{2x + k}{x + k}} \leqslant \frac{1}{\sqrt{kx}} \left[ \frac{2x}{k} - \frac{x}{k+x}\right]= \frac{1}{\sqrt{kx}} \frac{kx+ 2x^2}{k(k+x)}= \frac{k\sqrt{x} + 2 x^{3/2}}{k^{3/2}(k+x)}$$
For any fixed $x> 0$, the RHS is $\mathcal{O}(k^{-3/2})$ and the series is pointwise convergent by comparison with the convergent p-series $\sum k^{-3/2}$.
Can you finish now by showing the series is uniformly convergent on $(0,1)$ or any bounded interval $(0,c]$ using the inequality (2) and the Weierstrass M-test?