Show that $f: (0, \infty) \rightarrow (0, \infty)$ is a well-defined and continuous function, where $$f(x) = \sum_{n=0}^{\infty} \frac{1}{n(n+1)+x}.$$
I'm not exactly sure what it means for a function to be well-defined, but for continuity I was thinking of showing that $\{f_n\}$ converges uniformly by the Weierstrass M-test and then concluding that since $f(x)$ can be differentiated term by term, the derivative $f'(x)$ exists which implies continuity.
Does this work?
Just added for your curiosity.
$$f(x) = \sum_{n=0}^{\infty} \frac{1}{n(n+1)+x}=\frac{\pi \tan \left(\frac{\pi}{2} \sqrt{1-4 x}\right)}{\sqrt{1-4 x}}$$
In this case I think well-defined just means $f(x) \in (0,\infty)$.
Your argument about differentiation is unnecessary. Observe that
$$\sum_{n=1}^\infty \frac{1}{n(n+1) +x} \le \sum_{n=1}^\infty \frac{1}{n^2}$$ for all $x \in (0, \infty)$. So, the partial sums $\sum_{n=1}^N \frac{1}{n(n+1) +x}$ converge uniformly to $f$, and each is continuous. Now you can conclude.
