Let $n$ be a positive integer. Let $C \subseteq[0,1]$ be the ternary Cantor set. I have made the conjecture $C+\frac 1 nC=[0,1+1/n]$ and am wondering if it is true or not.
It is obviously true when $n=1$, since \begin{align*} C+C & = \left\{\frac {a_1+b_1} {3}+ \frac {a_2+b_2} {3^2}+...:a_i,b_i \in \{0,2\}\right\} \\ & = \left\{\frac {2c_1} 3+ \frac {2c_2} {3^2}+...: c_i\in\{0,1,2\}\right\} \\ & =2[0,1]=[0,2]. \end{align*}
Moreover, for general $n$, I see that $C+ \frac 1 n C$ has $1+1/n$ for its maximum and $0$ for its minimum. But how do we know that $C+ \frac 1 n$ contains everything in between? (Does it?)
For $n=2$, I see that $C+ \frac 1 2C=\{\frac {c_1} 3+ \frac {c_2} {3^2}+...:c_i \in \{0,1,2,3\}\}$. Intuitively, I want to "collapse" the the $c_i=3$ terms to the left but I'm having trouble making this rigorous.
I think it's true for $n = 2$ as well.
Certainly $\frac32 = 1 + \frac12 \in C + \frac{C}2$.
Suppose then that $x \in [0, \frac32).$
Define $s_0 = 0,$ $s_1 = 1,$ $s_2 = \frac43,$ and generally $$ s_k = 1 + \frac13 + \frac1{3^2} + \cdots + \frac1{3^{k-1}} \quad (k \geqslant 0). $$ Then $$ s_k = 3(s_{k+1} - 1) = u_k + v_k \quad (k \geqslant 0), $$ where \begin{align*} v_k = s_{k+1} - 1 & = 0.11\cdots1 \quad (k \text{ ternary digits}), \\ u_k = 2v_k & = 0.22\cdots2 \quad (k \text{ ternary digits}). \end{align*} There exists a unique non-negative integer $k$ such that $s_k \leqslant x < s_{k+1}.$
That is, $$ x = s_k + y, \text{ where } 0 \leqslant y < \frac1{3^k}. $$ Let the ternary expansion of $y$ be $$ y = 0.00\cdots0t_1t_2\cdots \quad (k \text{ zeros before } t_1). $$ Then $y = u^* + v^*$, where the ternary expansion of $u^*$ begins with $k$ zeros, and consists only of zeros and twos, and the ternary expansion of $v^*$ also begins with $k$ zeros, and consists only of zeros and ones.
Therefore $$ x = (u_k + u^*) + (v_k + v^*), $$ where $u_k + u^* \in C$, $v_k + v^* \in \frac{C}2.$
Summary
The conjecture is true for $n = 1, 2, 3,$ and false for $n > 3.$
As Wojowu noted in a comment on the question: if $n > 3,$ and $x, y \in C$, then $0 \leqslant y \leqslant 1,$ and either $x \geqslant \frac23,$ in which case $x + \frac{y}n \geqslant \frac23,$ or $x \leqslant \frac13,$ in which case $x + \frac{y}n \leqslant \frac13 + \frac1n.$ In both cases, $x + \frac{y}n \notin \left(\frac13+\frac1n, \frac23\right);$ so the set $C + \frac{C}n$ has no intersection with this interval.
Wojowu also noted that the case $n = 3$ of the conjecture follows from the case $n = 1.$ I don't know the exact proof (s)he had in mind, but here is one along similar lines: $3C = C \cup (C + 2),$ therefore $3C + C = (C + C) \cup (C + C + 2) = [0, 2] \cup [2, 4] = [0, 4],$ therefore $C + \frac{C}3 = \left[0, \frac43\right].$
As for the case $n = 1$, I'm afraid I can't follow the proof given in the question (it's probably just me!), so here is another proof (perhaps it's the same one): $C$ consists of all real numbers expressible as $\sum_{n = 1}^\infty t_n/3^n,$ where each ternary digit $t_n$ is 0 or 2, therefore $\frac{C}2$ consists of all infinite or finite sums (including the empty sum) of the form $\sum_k 3^{-n_k}$ ($0 < n_1 < n_2 < \dotsb$), therefore $$ \frac{C}2 + \frac{C}2 = \left\{ \sum_{n=1}^\infty\frac{t_n}{3^n}: t_n \in \{0,1,2\}\right\} = [0,1], $$ therefore $C + C = [0, 2].$