Proving $ \sum_{n=1}^{ \infty } f_n(1) $ converges when $ \sum_{n=1}^{ \infty } f_n(x) $ converges uniformly I have a question which I tried solving for a few hours.
My only solid direction so far is trying using Cauchy.
Let $(f_n(x))$ be a series of continuous functions in $[0,1]$.
Prove; If $ \sum_{n=1}^{ \infty }  f_n(x) $ converges uniformly at $[0,1)$, then $ \sum_{n=1}^{ \infty }  f_n(1) $ converges.
Thank you in advance, been trying a lot solving this. Much appreciated.
Maybe not the most elegant, but a detailed way to get it by &quot;splitting $\varepsilon$.&quot;
Let $S_n\colon[0,1]\to\mathbb{R}$ be the function defined by
$S_n(x) = \sum_{k=1}^n f_k(x)$, for $n\geq 1$ and $x\in[0,1]$. Since the $f_n$'s are continuous on a compact set, they are uniformly continuous, so is $S_n$. Moreover, by assumption the sequence $(S_n)_n$ converges uniformly on $(0,1)$, to a limit which we will call $S\colon(0,1)\to\mathbb{R}$.
Note that $S$ is then uniformly continuous (can you see why?), and thus can be extended on $[0,1]$ to a continuous function. I.e., we get that $S$ is continuous on $[0,1]$.

Fix $\varepsilon &gt; 0$. By uniform convergence, there exists $n_\varepsilon\geq 1$ such that, for every $n\geq n_\varepsilon$, $\sup_{x\in(0,1)} |S_n(x)-S(x)| \leq \varepsilon/3$.

Moreover, since $S_{n_\varepsilon}$ is continuous on $[0,1]$, there exists $\delta_\varepsilon&gt;0$ such that, for all $x\in(1-\delta_\varepsilon,1)$, $|S_{n_\varepsilon}(1)-S_{n_\varepsilon}(x)| \leq \varepsilon/3$. (Using continuity at $1$).

Moreover, since $S$ is also continuous on $[0,1]$, there exists $\delta'_\varepsilon&gt;0$ such that, for all $x\in(1-\delta'_\varepsilon,1)$, $|S(1)-S(x)| \leq \varepsilon/3$. (Using continuity at $1$).


By the triangle inequality, combining the two inequalities for $S_{n_\varepsilon}$, we get that for all $x\in(1-\min(\delta_\varepsilon,\delta'_\varepsilon),1)$,
$$
|S_{n_\varepsilon}(1)-S(1)| \leq \varepsilon
$$
Since $\varepsilon&gt;0$ was arbitrary, this shows that the sequence $(S_n(1))_n$ converges, which is what we wanted.
If $f(x)=\sum_{n=1}^\infty f_n(x)$ converges uniformly on $(0,1)$, then $\forall \epsilon&gt;0$ there exists $N$ s.t. $$\left|\sum_{n=N}^\infty f_n(x) \right| &lt; \epsilon$$ for all $x\in (0,1)$. Furthermore $\sum_{n=N}^\infty f_n(x)$ is uniformly continuous on $(0,1)$ (and can therefore be continued to $[0,1]$) since (by uniform continuity of $f_n(x)$ on $(0,1)$ and uniform convergence of $f(x)$) $\forall \epsilon'&gt;0$ there is $N'&gt;N$ and $\delta(N')=\delta'&gt;0$, s.t. $\forall x,x' \in (0,1)$ with $|x-x'|&lt;\delta'$ $$\left|\sum_{n=N}^\infty f_n(x)-f_n(x') \right| \leq \sum_{n=N}^{N'-1} \underbrace{\left| f_n(x)-f_n(x') \right|}_{&lt;\frac{\epsilon'}{N'-N}} + \underbrace{\left|\sum_{n=N'}^\infty f_n(x)\right|}_{&lt;\epsilon'} + \underbrace{\left|\sum_{n=N'}^\infty f_n(x') \right|}_{&lt;\epsilon'}&lt;3\epsilon' \, .$$
For $x=1-\delta$ with $\delta &gt;0$ this gives
$$\left|\sum_{n=N}^\infty f_n(1) \right| \leq \left|\sum_{n=N}^\infty f_n(1) - f_n(1-\delta) \right| + \left|\sum_{n=N}^\infty f_n(1-\delta) \right|&lt;2\epsilon$$
The second term is less than $\epsilon$ by definition of uniform convergence, while the first term can be made less than $\epsilon$ for small enough $\delta$ due to the continuity of $\left|\sum_{n=N}^\infty f_n(x) \right|$ on $[0,1]$.