Prove\disprove: $\sum_{k\ge1} |a_{n,k}| \sqrt{k} \le 1 \implies lim_{n \to \infty} \sum_{k\ge1} a_{n,k} = \sum_{k\ge1} lim_{n \to \infty} a_{n,k}$ Let $\{a_{n,k}, S_n\}$ be a set of real numbers such that $\sum_{k\ge1} a_{n,k} = S_n$ for every $n$, and say that $lim_{n \to \infty} S_n$ exists.
Prove or disprove the following statement:
$\sum_{k\ge1} |a_{n,k}| \sqrt{k} \le 1 \implies lim_{n \to \infty} \sum_{k\ge1} a_{n,k} = \sum_{k\ge1} lim_{n \to \infty} a_{n,k}$
I tend to believe it's true. I tried to use all sorts of tricks to prove it, including the Dominant Convergence Theorem, but it didn't succeed.
Does anyone have any idea?
 A: Yes! I did it eventually. My proof involves nothing but limits calculation, no dominant convergence theorem, no measure theory or anything like that. I do believe that there's a much shorter proof using more advanced tools, so if anyone has it I'd be very glad to see it.
The proof:
Let $\varepsilon>0$, and let $k_0$ be some integer $> \frac{1}{\varepsilon^2}$ 
Claim 1: for every $n \in \mathbb{N}$, $\sum_{k=k_0}^\infty |a_{n,k}| < \varepsilon$.
Suppose that there exists an $n$ such that $\sum_{k=k_0}^\infty |a_{n,k}| \ge \varepsilon$. Than you'd have: 
$$1 \ge \sum_{k=0}^\infty |a_{n,k}| \sqrt{k} \ge \sum_{k=k_0}^\infty |a_{n,k}| \sqrt{k} \ge \sqrt{k_0}\sum_{k=k_0}^\infty |a_{n,k}| > \frac{1}{\varepsilon}\sum_{k=k_0}^\infty |a_{n,k}| \ge \frac{1}{\varepsilon}\varepsilon = 1$$
And you've got $1>1$ which is of course false.
Claim 2: $\sum_{k=k_0}^\infty \lim \limits_{n \to \infty} |a_{n,k}| \le \varepsilon$
Since this is a positive series it's enough to show that $\sum_{k=k_0}^m \lim \limits_{n \to \infty} |a_{n,k}| \le \varepsilon$ for every $m$.
$$\sum_{k=k_0}^m \lim \limits_{n \to \infty} |a_{n,k}| = \lim \limits_{n \to \infty} \sum_{k=k_0}^m |a_{n,k}| \le \lim \limits_{n \to \infty} \sum_{k=k_0}^\infty |a_{n,k}| \le \lim \limits_{n \to \infty} \varepsilon = \varepsilon$$
Claim 3: $\lim \limits_{n \to \infty} \sum_{k=k_0}^\infty a_{n,k}$ exists, and also $|\lim \limits_{n \to \infty} \sum_{k=k_0}^\infty a_{n,k}| \le \varepsilon$
This is almost obvious since
$$\sum_{k=k_0}^\infty a_{n,k} = \sum_{k=1}^\infty a_{n,k} -\sum_{k=1}^{k_0-1} a_{n,k}$$
It is given that the first expression on the right side has a limit as $n \to \infty$, and the second expression is a finite sum of convergent sequence.
also, for every $n$:
$$|\sum_{k=k_0}^\infty a_{n,k}| \le \sum_{k=k_0}^\infty |a_{n,k}| < \varepsilon \implies \lim \limits_{n \to \infty} |\sum_{k=k_0}^\infty a_{n,k}| \le \varepsilon $$
And now the rest of the proof is easy.
$$|\sum_{k=1}^\infty \lim \limits_{n \to \infty} a_{n,k} -\lim \limits_{n \to \infty} \sum_{k=1}^\infty a_{n,k}| =$$
 $$=|\sum_{k=1}^{k_0-1} \lim \limits_{n \to \infty} a_{n,k} + \sum_{k=k_0}^\infty \lim \limits_{n \to \infty} a_{n,k} -\lim \limits_{n \to \infty} \sum_{k=1}^{k_0-1} a_{n,k} -\lim \limits_{n \to \infty} \sum_{k=k_0}^\infty a_{n,k}|$$
The first and third expressions cancel each-other, since the limit of finite sum is the sum of the limits, and we stay with:
$$|\sum_{k=k_0}^\infty \lim \limits_{n \to \infty} a_{n,k} -\lim \limits_{n \to \infty} \sum_{k=k_0}^\infty a_{n,k}| \le |\sum_{k=k_0}^\infty \lim \limits_{n \to \infty} a_{n,k}| + |\lim \limits_{n \to \infty} \sum_{k=k_0}^\infty a_{n,k}| \le$$
$$\le \sum_{k=k_0}^\infty \lim \limits_{n \to \infty} |a_{n,k}| + \varepsilon \le 2\varepsilon$$
and since $\varepsilon$ is arbitrary, we get 
$$\sum_{k=1}^\infty \lim \limits_{n \to \infty} a_{n,k} = \lim \limits_{n \to \infty} \sum_{k=1}^\infty a_{n,k}$$
$Q.E.D$
