Limit at infinity of a function series 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.
 A: Let $\;\mathbb{N}_{0}=\mathbb{N}\cup\{0\}=\left\{0,1,2,3,4,5,\ldots\right\}\;.$
Theorem:
If $\;\big\{f_n:\left[a,+\infty\right[\to\mathbb{R}\big\}_{n\in\mathbb{N}_{0}}\;$ is a sequence of functions ,
$\sum\limits_{n=0}^\infty f_n(x)\;$ is an uniformly convergent series on $\;\left[a,+\infty\right[\;$ to a function $\;f:\left[a,+\infty\right[\to\mathbb{R}\;,$
$\lim\limits_{x\to+\infty} f_n(x)=l_n\in\mathbb{R}\;\;\;$ for all $\;n\in\mathbb{N}_{0}\;,$
and $\;\sum\limits_{n=0}^\infty l_n=l\in\overline{\mathbb{R}}=\mathbb{R}\cup\{-\infty,+\infty\}\;,$
then $\;\lim\limits_{x\to+\infty} f(x)=l\;.$
Proof:
I am going to prove the theorem in the case that $\;l\in\mathbb{R}\;.$
Let $\;\epsilon\;$ be any positive real number $\;(\epsilon>0)\;.$
Since $\;\sum\limits_{n=0}^\infty f_n(x)\;$ is a uniformly convergent series on $\;\left[a,+\infty\right[\;,\;$ there exists $\;\nu_\epsilon\in\mathbb{N}\;$ such that $\;\forall n>\nu_\epsilon\;$ and $\;\forall x\in\left[a,+\infty\right[\;$ it results that $\;\left|\sum\limits_{k=0}^n f_k(x)-f(x)\right|<\cfrac{\epsilon}{3}\;.\quad\color{blue}{(1)}$
Since $\;\sum\limits_{n=0}^\infty l_n=l\in\mathbb{R}\;,\;$ there exists $\;\mu_\epsilon\in\mathbb{N}\;$ such that $\;\forall n>\mu_\epsilon\;$ it results that $\;\left|\sum\limits_{k=0}^n l_k-l\right|<\cfrac{\epsilon}{3}\;.\quad\color{blue}{(2)}$
Let $\;\overline{n}_\epsilon=\max\{\nu_\epsilon,\mu_\epsilon\}+1\in\mathbb{N}\;.\;$
It implies that $\;\overline{n}_\epsilon>\nu_\epsilon\;$ and $\;\overline{n}_\epsilon>\mu_\epsilon\;.$
Since $\lim\limits_{x\to+\infty} f_n(x)=l_n\in\mathbb{R}\;\;$ for all $\;n\in\mathbb{N}_{0}\;,\;$ by applying the Limit Sum Law, it follows that
$\lim\limits_{x\to+\infty}\sum\limits_{k=0}^{\overline{n}_\epsilon}f_k(x)=\sum\limits_{k=0}^{\overline{n}_\epsilon}l_k\;,$
so there exists $\;b_\epsilon\ge a\;$ such that $\;\forall x\in\left]b_\epsilon,+\infty\right[\;$ it results that $\;\left|\sum\limits_{k=0}^{\overline{n}_\epsilon}f_k(x)-\sum\limits_{k=0}^{\overline{n}_\epsilon}l_k\right|<\cfrac{\epsilon}{3}\;.\quad\color{blue}{(3)}$
Moreover, for all $\;x\in\left]b_\epsilon,+\infty\right[\;$ it results that
$\big|f(x)-l\big|=$
$=\left|f(x)-\sum\limits_{k=0}^{\overline{n}_\epsilon}f_k(x)+\sum\limits_{k=0}^{\overline{n}_\epsilon}f_k(x)-\sum\limits_{k=0}^{\overline{n}_\epsilon}l_k+\sum\limits_{k=0}^{\overline{n}_\epsilon}l_k-l\right|\le$
$\le\left|\sum\limits_{k=0}^{\overline{n}_\epsilon}f_k(x)-f(x)\right|+\left|\sum\limits_{k=0}^{\overline{n}_\epsilon}f_k(x)-\sum\limits_{k=0}^{\overline{n}_\epsilon}l_k\right|+\left|\sum\limits_{k=0}^{\overline{n}_\epsilon}l_k-l\right|<$
$\underset{\overbrace{\text{from (1), (3) and (2)}}}{<}\cfrac{\epsilon}{3}+\cfrac{\epsilon}{3}+\cfrac{\epsilon}{3}=\epsilon\;.$
Therefore,
$\forall \epsilon>0\;\;\exists\;b_\epsilon\ge a\;$ such that $\;\forall x\in\left]b_\epsilon,+\infty\right[\;$ it results that $\big|f(x)-l\big|<\epsilon\;,\;$ that is to say ,
$\lim\limits_{x\to+\infty} f(x)=l\;.$
Now I am going to prove the theorem in the case that $\;l=+\infty\;.$
Let $\;M\;$ be any positive real number $\;(M>0)\;.$
Since $\;\sum\limits_{n=0}^\infty f_n(x)\;$ is a uniformly convergent series on $\;\left[a,+\infty\right[\;,\;$ there exists $\;\nu_\epsilon\in\mathbb{N}\;$ such that $\;\forall n>\nu_\epsilon\;$ and $\;\forall x\in\left[a,+\infty\right[\;$ it results that $\;\left|\sum\limits_{k=0}^n f_k(x)-f(x)\right|<M\;.\quad\color{blue}{(4)}$
Since $\;\sum\limits_{n=0}^\infty l_n=l=+\infty\;,\;$ there exists $\;\mu_\epsilon\in\mathbb{N}\;$ such that $\;\forall n>\mu_\epsilon\;$ it results that $\;\sum\limits_{k=0}^n l_k>3M\;.\quad\color{blue}{(5)}$
Let $\;\overline{n}_\epsilon=\max\{\nu_\epsilon,\mu_\epsilon\}+1\in\mathbb{N}\;.\;$
It implies that $\;\overline{n}_\epsilon>\nu_\epsilon\;$ and $\;\overline{n}_\epsilon>\mu_\epsilon\;.$
Since $\lim\limits_{x\to+\infty} f_n(x)=l_n\in\mathbb{R}\;\;$ for all $\;n\in\mathbb{N}_{0}\;,\;$ by applying the Limit Sum Law, it follows that
$\lim\limits_{x\to+\infty}\sum\limits_{k=0}^{\overline{n}_\epsilon}f_k(x)=\sum\limits_{k=0}^{\overline{n}_\epsilon}l_k\;,$
so there exists $\;b_\epsilon\ge a\;$ such that $\;\forall x\in\left]b_\epsilon,+\infty\right[\;$ it results that $\;\left|\sum\limits_{k=0}^{\overline{n}_\epsilon}f_k(x)-\sum\limits_{k=0}^{\overline{n}_\epsilon}l_k\right|<M\;.\quad\color{blue}{(6)}$
Moreover, for all $\;x\in\left]b_\epsilon,+\infty\right[\;$ it results that
$f(x)=f(x)-\sum\limits_{k=0}^{\overline{n}_\epsilon}f_k(x)+\sum\limits_{k=0}^{\overline{n}_\epsilon}f_k(x)-\sum\limits_{k=0}^{\overline{n}_\epsilon}l_k+\sum\limits_{k=0}^{\overline{n}_\epsilon}l_k\ge$
$\ge-\left|\sum\limits_{k=0}^{\overline{n}_\epsilon}f_k(x)-f(x)\right|-\left|\sum\limits_{k=0}^{\overline{n}_\epsilon}f_k(x)-\sum\limits_{k=0}^{\overline{n}_\epsilon}l_k\right|+\sum\limits_{k=0}^{\overline{n}_\epsilon}l_k>$
$\underset{\overbrace{\text{from (4), (6) and (5)}}}{>}-M-M+3M=M\;.$
Therefore,
$\forall M>0\;\;\exists\;b_\epsilon\ge a\;$ such that $\;\forall x\in\left]b_\epsilon,+\infty\right[\;$ it results that $f(x)>M\;,\;$ that is to say ,
$\lim\limits_{x\to+\infty} f(x)=+\infty=l\;.$
Analogously it is possible to prove the theorem in the case that $\;l=-\infty\;.$

It is possible to prove the theorem by substituting uniformly convergence for another hypothesis?
I hope you will answer my question because I am very interested in it.
