Show that $\lim_{x \to 1} \sum_{n=1}^{\infty}\frac{n x^2}{n^3 + x^2}= \sum_{n=1}^{\infty} \frac{n}{n^3 +1}$. What I understand from the question is that the author (N. L. Carothers) what me to show that the series $\sum_{n=1}^{\infty}\frac{n x^2}{n^3 + x^2}$ converges uniformly. Hence, the exchange between the limit and the summation is possible. However, It's been a week that I have been trying to find a summable upper bound so that the M-test will be applicable with no improvements. I tried AM-GM but did not get any results. Honestly, I do not publish the post here unless I try everything I know.
 A: It is not necessary to prove uniform convergence. The series on both sides of the equation are convergent so we can consider the differenece and show that it goes to $0$. The differnce is $(x^{2}-1)\sum \frac {n^{4}} {(n^{3}+x^{2}) (n^{3}+1)}$ which tends to $0$ since  $0 \leq\sum \frac {n^{4}} {(n^{3}+x^{2}) (n^{3}+1)} \leq \sum\frac  1{n^{2}}$.
The series does not converge uniformly on the real line since the general term does not tend to $0$ uniformly. [ Choose $x$ such that $\frac {nx^{2}} {n^{3}+x^{2}} =1$ to see this]. However, the series converges uniformly on $[-L,L]$ for any $L$: Check that  $\frac {nx^{2}} {n^{3}+x^{2}} \leq \frac {L^{2}} {n^{2}}$ for $|x| \leq L$. Of course,  this is good enough for proving the result.
A: Since you want to find the limit of the series on the left as $x$ goes to 1. You can prove the convergence in any neighborhood of 1.
Consider any neighborhood of 1, namely, consider the neigborhood $(1-\delta, 1+ \delta)$ where $\delta >0$. So
$\lVert \frac{n x^2}{n^3+x^2}\rVert_\infty = \sup_{x\in (1-\delta,1+\delta) } \big|\frac{n x^2}{n^3+x^2} \big| \le \big|\frac{x^2}{n^2}\big| \le \big| \frac{1+\delta}{n^2} \big|.$
Now, we found an upper bound $M_n = \frac{1+\delta}{n^2}$ and according to the M-test theorem $\sum_{n=1}^{\infty} \frac{1+\delta}{n^2}$ converges on the negoborhood $|x-1| <\delta,\,\,\, \forall \delta >0$.
