Does this sequence of function's sum uniformly converge? $Q)$ For the countable set, $\mathbb{Q}$ (rational number's set)
$$(0,1) \cap \mathbb{Q} = \{a_1, a_2, \ldots, a_n ,\ldots\}$$ 
Define $f_n : [0,1] \to \mathbb{R}$ by $\begin{cases}
1/n,  & x=a_n \\
0 & x \neq a_n
\end{cases}$
Determine uniformly converge $\sum_{n=1}^\infty f_n(x)$  on $[0,1]$

My guess this is not uniformly converge 
since  $a_n \in (0,1)$ into $f_n(x)$
Then, $\sum_{n=1}^{\infty} f_n(a_n)=\sum_{n=1}^{\infty}{ 1 \over {n}} $ 
Therefore,  $\sum_{n=1}^\infty f_n(x)$ does not uniformly converge.
What do you think? Any help would be appreciated.
 A: It does converge uniformly if $a_1,a_2,a_3,\ldots$ are all distinct.
For example suppose we have
$$
\begin{array}{lcll}
a_1 & = & 1/2 & (\text{denominator} = 2) \\
a_2, a_3 & = & 1/3,\,\, 2/3 & (\text{denominator} = 3) \\
a_4, a_5 & = & 1/4, \, \,3/4 & (\text{denominator} = 4) \\
a_6,a_7,a_8,a_9 & = & 1/5,\,\, 2/5,\,\, 3/5,\,\, 4/5 & (\text{denominator} = 5) \\
a_{10}, a_{11} & = & 1/6,\,\,5/6 & (\text{denominator} = 6) \\
\text{etc.}
\end{array}
$$
Then
$$
\sum_{n=1}^{11} f_n(x) =
\begin{cases}
1 & \text{if } x = 1/2, \\
1/2 & \text{if } x=1/3, \\
1/3 & \text{if } x=2/3, \\
1/4 & \text{if } x = 1/4, \\
1/5 & \text{if } x = 3/4, \\
1/6 & \text{if } x = 1/5, \\
1/7 & \text{if } x = 2/5, \\
1/8 & \text{if } x = 3/5, \\
1/9 & \text{if } x = 4/5, \\
1/10 & \text{if } x = 1/6, \\
1/11 & \text{if } x = 5/6, \\
0 & \text{if } x = \text{anything else.}
\end{cases}
$$
At no point do we have $1+\frac 1 2 + \frac 1 3 + \cdots + \frac 1 {11}.$
This sum of $11$ terms differs from the sum of infinitely many terms by $1/12$ if $x=a_{12}$ and by less than that if $x={}$any other number. The biggest discrepancy between the sum of infinitely many terms and the sum of finitely many is always $1/(n+1)$ when $n$ terms have been added.
