A property about uniform convergence Suppose a function series $\sum_{n=1}^{\infty}u_n(x)$ converges on  $(a,b)$ and every $u_n(x)$ is continuous on the interval $(a,b)$,but this series diverges at $x=a$ or $x=b$, can we deduce that  $\sum_{n=1}^{\infty}u_n(x)$ does not converge uniformly on $(a,b)$?
For example, $\sum_{n=1}^{\infty}x^n$ converges on the interval $(0,1)$, but not uniformly on the interval, for it diverges at $x=1$.
I conjecture it is true,but I am struggling to give the proof.
Any correction or improvement is welcomed! Thanks in advance.

Consider, $u_n(x) = 0$ in $(a, b)$, $u_n(x) = 1, x=a$ or $x = b$. You get a uniform convergence in $(a, b)$ but divergence at the endpoints.
Continuity Edit:
Let $s_n$ be the partial sum of $u_i$'s up to term n.
Suppose, that the sequence does not converge at $a$, that is, 
$$\exists\epsilon > 0, \forall n_0 \exists n,m > n_0, |s_n(a) - s_m(a)| > \epsilon \tag{*} \label{*}$$
furthermore, suppose that,
$$ \forall n\forall\epsilon'> 0 \ \exists\delta > 0\ | x\in(a, a + \delta) \implies |s_n(x) - s_n(a)| \leq \epsilon' \tag{**} \label{**}$$
We will show that the sequence $s_n$ is not uniformly Cauchy.
Take $\epsilon$ given by $\eqref{*}$. Pick some $n_0$, take the given $n, m$ by $\eqref{*}$. For each of $n, m$ invoke $\eqref{**}$ with $\epsilon' = \epsilon / 4$. Take the minimum of the $\delta$'s given by \eqref{**} and an $x$ in that interval. Then, 
$$ |s_n(x) - s_m(x)| \geq |s_n(a) - s_m(a)| - |s_n(x) - s_n(a)| - |s_m(x) - s_m(a)| \geq \epsilon / 2$$
Therefore, 
$$ \exists\epsilon'' > 0, \forall n_0 \exists n,m > n_0, \exists x\in (a,b), |s_n(x) - s_m(x)| > \epsilon''$$
That is, the sequence is not uniformly cauchy.
The rest is showing that a sequence is uniformly cauchy iff it is uniformly convergent.
So the statement seems correct.
