In my researches I got stuck on two similar calculations, and I'd like to deal with them in one fell swoop.

1. I want to say that $$ \lim_{x \to \infty} \sum_{n > 1} z_n \!\!\!\sum_{\substack{d \mid n \\ 1 < d \le x}} \!\!\!\mu(d) = \sum_{n > 1} z_n \lim_{x \to \infty} \!\!\! \sum_{\substack{d \mid n \\ 1 < d \le x}} \!\!\!\mu(d) = \sum_{n > 1} z_n \sum_{\substack{d \mid n \\ 1 < d}} \mu(d) = -\!\sum_{n > 1} z_n$$ where the $z_n$ are complex. Here $\mu$ is the Mobius function, which satisfies the identity $\sum_{d \mid n} \mu(d) = 0$ for $n > 1$, whence the minus sign.

2. I want to say that $$ \lim_{s \to \infty} 2^s \zeta'(s) = -\!\lim_{s \to \infty} \sum_{n \ge 1} \log n \Big(\frac{2}{n}\Big)^s = -\!\log 2 -\!\sum_{n > 2} \log n \lim_{s \to \infty} \Big(\frac{2}{n}\Big)^s = -\!\log 2.$$ Here, of course, $\zeta$ is Riemann's zeta function.

The question is therefore this: there's a sequence of functions $f_n(x)$ with finite limits $a_n$ as $x \to \infty$. Under what conditions on the $f_n$ and the $a_n$ is it true that $$ \lim_{x \to \infty} \sum_n f_n(x) = \sum_n a_n? $$ Plainly, the series of limits should be summable; does it have to be absolutely convergent? Likewise, the series of functions should converge pointwise; does it have to converge absolutely? Uniformly?

Note that the function series in the second example is not uniformly convergent in a neighbourhood of infinity (I believe).

Counterexamples and links to other posts are all very welcome! I will accept an answer which remarks on my specific two examples if nothing useful can be said in the general case.

  • $\begingroup$ If everything converges absolutely then it is obvious you can invert $\lim$ and $\sum$. If it doesn't, you have to prove that $\lim_{N \to \infty}\lim_{x \to \infty} \sum_{n > N} \ldots \to 0$ and $ \lim_{x \to \infty}\lim_{N \to \infty} \sum_{n > N} \ldots \to 0$ $\endgroup$ – reuns Jul 21 '16 at 15:01
  • $\begingroup$ @user1952009 Good point! $\endgroup$ – Unit Jul 21 '16 at 18:09

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