If a series converges uniformly on $[a+\varepsilon,b) $ for all $\varepsilon>0$ and pointwise on $[a,b)$, is then the convergence uniform on $[a,b) $? The question arises from the specific series $\sum_{k=0}^\infty \dfrac{k (ex)^k}{(k+1)^2} $, which converges pointwise on $[-1/e,1/e) $ and uniformly on $[-1/e,1/e-\varepsilon] $ for every $\varepsilon&gt;0$. 

I found the intwrval of pointwise convergengence and I proved uniform convergence on $[-1/e+\varepsilon,1/e-\varepsilon] $, is this enough to conclude the interval of uniform convergence is $[-1/e,1/e-\varepsilon] $?
In general the above statement is false.
Consider, for example, $\sum_{n=0}^\infty x^n$, that converges pointwise to $f(x) = \frac{1}{1-x}$ for $x\in (-1,1)$ and uniformly on every compact interval $[a,b]\subset (-1,1)$.

On the other hand, you have $s_n(x) := \sum_{k=0}^n x^k = \frac{1-x^{n+1}}{1-x}$, and this sequence does not converge uniformly to $f$ in $(-1,1)$.
In general the above statement is false.
Consider, for example, $\sum_{n=0}^\infty x^n$, that converges pointwise to $f(x) = \frac{1}{1-x}$ for $x\in (-1,1)$ and uniformly on every compact interval $[a,b]\subset (-1,1)$.

On the other hand, you have $s_n(x) := \sum_{k=0}^n x^k = \frac{1-x^{n+1}}{1-x}$, and this sequence does not converge uniformly to $f$ in $(-1,1)$.