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I am trying to show the following:

$\lim_{x \rightarrow 1}\sum{\frac{nx^2}{n^4+x^2}} = \sum{\frac{n}{n^4+1}}$, and I'd like to do it using ONLY THE THEOREMS AND TOOLS PROVIDED IN CLASS SO FAR. Specifically, my class has covered up through chapter 7 in Baby Rudin. Please don't recommend proofs not specifically covered in that text, or prove additional theorems to solve this problem; it is NOT what I'm looking for. Something similar was asked here, but it doesn't address my question: Moving a limit inside an infinite sum

Any, I'm mostly done, and have proceeded thus:

1) Essentially, this is asking us to show that the series converges uniformly, which I know heuristically means I can move the limit inside the sigma (more on this later). This I have done using Theorem 7.10 in Rudin, which in briefs states that if you can bounds each $f_n(x)$ with some $M_n$ (with no $x$ dependence), such that $\sum{M_n}$ converges, then the sum $\sum{f_n(x)}$ converges uniformly. So far so good

2) Next, I show (which was more difficult) that $\lim {x\rightarrow 1} \frac{nx^2}{n^4+x^2} = \frac{n}{n^4 + 1}$. This took some finessing but I'm satisfied with my work.

But I really need help with 3), below:

3). Finally, I need to justify the actual movement inside the summation. Rudin doesn't specifically refer to such movement in the text. What he does say is the following:

If $f_n(x) \rightarrow f(x)$ uniformly, then $\lim_{t \rightarrow x}\lim_{n \rightarrow \infty}{f_n(t)}= \lim_{n \rightarrow \infty} \lim_{t \rightarrow x}{f_n(t)}$

My thought is that if I let each $k_{th}$ partial sum equal $f_k(x)$, then the infinite sum is $\sum{k \rightarrow \infty}{f_k(x)}$, then I can claim that:

$\lim_{x \rightarrow 1}\lim_{k \rightarrow \infty}{f_k(t)}= \lim_{k \rightarrow \infty} \lim_{x \rightarrow 1}{f_k(t)}$, but each $f_k$ is itself a (finite) sum. Is it totally trivial to pass the limit inside the sum since (barring convergence issues) the sum of the limits is the limit of the sums?

BenL
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  • If $\sum_{n\geq 1}f_n(x)$ is a series of continuous functions, uniformly converging to some $f(x)$ on the interval $[0,2]$, then $$\lim_{x\to 1}\sum_{n\geq 1}f_n(x) = \sum_{n\geq 1}f_n(1) $$ by the continuity of $\sum_{n\geq 1}f_n(x)$ and uniform convergence. That is quite trivial. – Jack D'Aurizio Apr 26 '17 at 19:19
  • If it isn't to you, just prove this fact as an exercise. – Jack D'Aurizio Apr 26 '17 at 19:20
  • That's what I attempt to do in my post, using Rudin's actual theorems. If you're not going to actually read my post and answer my question, then I'm not clear why you're bothering to reply at all. – BenL Apr 26 '17 at 19:21
  • Rudin's way or not, that is trivial. The only difficult part of this exercise is to prove that we have uniform convergence over, say, $[0,2]$. If you really got that, you do not need any help. Voting for closing. – Jack D'Aurizio Apr 26 '17 at 19:23
  • The only theorem Rudin provides is that uniform convergence of sequences allows us to switch the limits. Series are never mentioned, and obviously it is NOT trivial to me because I bothered to spend time typing up a question. – BenL Apr 26 '17 at 19:26
  • Isn't a series a limit of a sequence, BenL? – Jack D'Aurizio Apr 26 '17 at 19:27
  • If you know conditions that allow you to switch limits, you also know conditions that allow you to switch $\lim_{x\to 1}$ and $\sum_{n\geq 1}$, don't you? – Jack D'Aurizio Apr 26 '17 at 19:28
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    I don't know Jack D'Aurizio, did you actually read my post? And no, I don't know anything. I wrote something that I think is true. You ignored my actual question and proceeded to lecture me about triviality. I can't even figure out whom I should complain to about how impolite and offensive you've been about this question. Obviously these are not trivial to me, or I wouldn't have asked. – BenL Apr 26 '17 at 19:29
  • I do not want to sound rude, but you should probably study Rudin's book a bit better before asking how to perform trivial manipulations / how to prove known theorems, clearly stated by Rudin. First things first: you started this post by writing that it looks easy to prove uniform convergence, in this case: well, it isn't. So, before digging into further details, you should probably explain how you really tackled your point $(1)$, since $\sum_{n\geq 1}\sup_{x\in\mathbb{R}^+}\frac{n x^2}{n^4+x^2}$ is not convergent. – Jack D'Aurizio Apr 26 '17 at 19:33
  • Let us continue this discussion in chat. – BenL Apr 26 '17 at 19:35
  • What is really easy is to conclude from uniform convergence, since $\lim_{x\to 1}\sum_{n=1}^{N}f_n(x) = \sum_{n=1}^{N}f_n(1)$ is an arbitrarily good approximation of $\lim_{x\to 1}\sum_{n\geq 1}f_n(x)$, and $\sum_{n=1}^{N}f_n(1)$ converges to $\sum_{n\geq 1}f_n(1)$ as $N\to +\infty$. – Jack D'Aurizio Apr 26 '17 at 19:36

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