# prove $(\sum_{n=0}^k f_n(x))_{k \in \mathbb{N}}$ converges uniformly over $X$ where each $f_n$ is bounded

Let $(f_n)_n$ be a sequences of from $X$ to $\mathbb{R}$ such that $\forall n \in \mathbb{N}$ $\exists M_n > 0$ such that $\sup_X \lvert f_n \rvert \le M_n$ and $\sum_{n=0}^{\infty} M_n < \infty$. Prove $(\sum_{n=0}^k f_n(x))_{k \in \mathbb{N}}$ converges uniformly.

The only thing I can think of is somehow using Cauchy but I have not had luck. I have been using the inequality given but all I seem to be getting is a bound for the sequence which does not guarantee convergence.

I have a theorem in my notes that states:

if a sequence of functions from some set to a complete metric space is uniformly Cauchy then there exists a function such that the sequence converges uniformly to. So it seems if I can get that this sequence is uniformly Cauchy over $X$ then I would have it.

So it would be $\forall \epsilon > 0 , \exists N, \forall p,q \ge N, sup_X \lvert f_p(x) - f_q(x) \rvert < \epsilon$.

If the sequence is monotonically increasing then using the boundedness I get that it has a Cauchy subsequence.

Edit: I forgot about Bolzano-Weierstrass. It is a bounded sequence so it has a convergent subsequence.

$$\left|\sum_{n=0}^k f_n(x) - \sum_{n=0}^j f_n(x)\right| = \left|\sum_{n=j+1}^k f_n(x)\right| \leqslant \sum_{n=j+1}^k |f_n(x)| \leqslant \ldots$$
• $$\left|\sum_{n=0}^k f_n(x) - \sum_{n=0}^j f_n(x)\right| = \left|\sum_{n=j+1}^k f_n(x)\right| \leqslant \sum_{n=j+1}^k |f_n(x)| \leqslant \sum_{n=j+1}^k M_n < \infty$$. The idea that I have is that $\sum_{n+0}^\infty M_n$ is monotone increasing and bounded so it gives a sequence which is Cauchy. So by the above inequalities we get the sequence of functions is also Cauchy. Feb 28 '17 at 5:20
• @oliverjones: .. or just that $\sum M_n$ is convergent. Very good!