Uniform convergence $\sum _{n=0} ^{\infty} \frac{\log (1+nx)}{nx^n}$ on $x \in (1,\infty)$

Question

How could I prove that $\sum _{n=0} ^{\infty} \frac{\log (1+nx)}{nx^n}$ is not uniformly convergent on an interval $I=(1,+\infty)$?
So far I have been thinking of proving that $f_n(x)=\frac{\log (1+nx)}{nx^n}$ does not uniformly converge on the interval, which leads me to finding $\lim _{n\to \infty}\sup_{x>1} \frac{\log (1+nx)}{nx^n}$. But there could be a nicer way.

Answer 1

It's easy to see that each summand in the series is bounded on $(1,\infty).$ If the series were uniformly convergent on $(1,\infty),$ then the series would sum to a bounded function there.

To show this fails, note that

$$\ln(1+nx) > \ln (nx) = \ln n + \ln x > \ln n.$$

Thus our series is bounded below by

$$\tag 1\sum_{n=1}^{\infty}\frac{\ln n}{nx^n}.$$

We're done if we show that as $x\to 1^+,$ $(1)\to \infty.$ That is easy to do, since $\sum (\ln n)/n=\infty.$