Uniformly convergence of $\sum_{n = 1}^{\infty} \frac{\sqrt{nx}}{x + n} \; \ln{ \left( 1 + \frac{x}{n} \right) }, \;\;\; x > 1$

Question

I want to investigate the uniform convergence of these series:

$$\sum_{n = 1}^{\infty} \frac{\sqrt{nx}}{x + n} \; \ln{ \left( 1 + \frac{x}{n} \right) }, \;\;\; x > 1$$

I think that the series converges, but I don't know how to prove it. Help me, please

Answer 1

$$S_N=\sum_{n = 1}^{N} \frac{\sqrt{nx}}{x + n} \; \ln{ \left( 1 + \frac{x}{n} \right) }, \;\;\; x > 1$$

$$\begin{align}\sup_{x}|S-S_N|&=\sup_{x}\sum_{n = N+1}^{\infty} \frac{\sqrt{nx}}{x + n} \; \ln{ \left( 1 + \frac{x}{n} \right) }\\ \\ &\ge \sum_{n = N+1}^{\infty} \frac{\sqrt{n(N+1)}}{(N+1) + n} \; \ln{ \left( 1 + \frac{(N+1)}{n} \right) }\\ \\ &\ge\frac{\sqrt{(N+1)(N+1)}}{(N+1) + (N+1)} \; \ln{ \left( 1 + \frac{(N+1)}{(N+1)} \right) }=2\ln2\end{align}$$

Therefore,

$$\limsup |S-S_N|\ge 2\ln2>0$$

It is not uniformly convergent.