I have some problems with internet connection, so do not be surprised if I stop talking suddenly. Anyway...

I think that inclusion you've mentioned shows $\sup_{k\ge n}(a_k+b_k) \le \sup_{j,k\ge n}(a_k+b_j)$. If you want to go into so much detail (i.e., if you do not take $\sup (a_k+b_k) \le \sup a_k + \sup b_k$ as obvious), you should probably explain also how $\sup_{j,k\ge n}(a_k+b_j) \le \sup_{k\ge n} a_k + \sup_{j\ge n} b_j$.

In fact, maybe you could show $\sup (a_k+b_k) \le \sup a_k + \sup b_k$ directly, without this intermediate step.

@user193319 Other than that it looks ok to me.

math.stackexchange.com/… Any clues?

Link not visible

@MartinSleziak Ah, I see. Thanks for pointing those out! By the way, would you happen to know the answer to my question about the excerpt from Conway's Complex Analysis, the one just above my last post?

@user193319 Isn't this actually Theorem 1.3 in Conway's Functions of one complex variable?

> (c) If $0<r<R$ then the series converges uniformly on $\{z; |z|\le r\}$.

> The number $R$ is called the radius of convergence of the power series.

@MartinSleziak Oh, and to get uniform convergence on any compact set, just use that fact compact sets are closed and bounded and can therefore be fit into some closed $r$-ball at the origin?

Looking at the proof, Weierstrass M-test is used to show what happens inside radius of convergence.

@user193319 Yes, if we are dealing with subsets of $\mathbb R$ or $\mathbb C$ (like here), we can use this characterization of compactness.

BTW you have asked a few questions related to limit superior/inferior. I'd guess at least some of them can be found if you look on the frequent tab of the limsup-and-liminf tag.

Oh, this is nice! Thank you for the link!

In any case, I think that knowing that those three or four definitions of limsup/liminf that often appear are equivalent is quite useful before trying to prove some other stuff about them.