We all know that $\sum_{k=0}^n\binom{n}k$ is not divisible by $3$.
QUESTION. Is it true that the numerator of $a_n$ (in reduced form) is never divisible by $3$? $$a_n=\sum_{k=0}^n\frac1{\binom{n}k}.$$
POSTSCRIPT. This the most detailed and pedagogical answer to any question that I asked here on MO. Thank you, Darij Grinberg.
Here is the answer I hinted at in a comment, in real detail. Took me a while, but I had no idea how tiresome such arguments are to expose...
Yes, it is true: see Corollary 6 (b) below. The proof relies on the concept of a $p$-adic valuation of a rational number. It is defined as follows:
Definition. Let $p$ be a prime. For each rational number $r$, we shall now define an element $v_{p}\left( r\right) $ of the set $\mathbb{Z}\cup\left\{ \infty\right\} $ (where $\infty$ is understood to be a symbol that satisfies the rules $\infty\geq n$ and $\infty+n=\infty$ for all $n\in\mathbb{Z} \cup\left\{ \infty\right\} $). We define it in three steps:
If $r=0$, then we set $v_{p}\left( r\right) =\infty$.
If $r\in\mathbb{Z}$, then we let $v_{p}\left( r\right) $ be the largest $k\in\mathbb{N}$ satisfying $p^{k}\mid r$. (Here and in the following, $\mathbb{N}$ means the set $\left\{ 0,1,2,\ldots\right\} $.)
If $r$ is any nonzero rational number, then we define $v_{p}\left( r\right) $ by $v_{p}\left( r\right) =v_{p}\left( a\right) -v_{p}\left( b\right) $, where $a$ and $b$ are two nonzero integers satisfying $r=a/b$. (This is well-defined, since any nonzero rational number $r$ can be written as $r=a/b$ for two nonzero integers $a$ and $b$, and since $v_{p}\left( a\right) -v_{p}\left( b\right) $ does not depend on the specific choice of $a$ and $b$. Also, this new definition of $v_{p}\left( r\right) $ for rational numbers $r$ does not conflict with the previous definition of $v_{p}\left( r\right) $ for integers $r$. This is all easy to check.)
Thus, an element $v_{p}\left( r\right) \in\mathbb{Z}\cup\left\{ \infty\right\} $ has been defined for each $r\in\mathbb{Q}$. This element $v_{p}\left( r\right) $ will be called the $p$-adic valuation of $r$.
We will need the following rules for $p$-adic valuations:
Proposition 1. Let $p$ be a prime.
(a) We have $v_{p}\left( ab\right) =v_{p}\left( a\right) +v_{p}\left( b\right) $ for any $a,b\in\mathbb{Q}$.
(b) We have $v_{p}\left( a+b\right) \geq\min\left\{ v_{p}\left( a\right) ,v_{p}\left( b\right) \right\} $ for any $a,b\in\mathbb{Q}$.
(c) We have $v_{p}\left( a^{k}\right) =kv_{p}\left( a\right) $ for any $a\in\mathbb{Q}$ and $k\in\mathbb{N}$.
(d) For any $i\in\mathbb{N}$ and $n\in\mathbb{Z}$, we have the equivalence $\left( p^{i}\mid n\right) \ \Longleftrightarrow\ \left( v_{p}\left( n\right) \geq i\right) $.
(e) If $a,b\in\mathbb{Q}$ satisfy $v_{p}\left( a\right) >v_{p}\left( b\right) $, then $v_{p}\left( a+b\right) =v_{p}\left( b\right) $.
(f) Let $s$ and $t$ be two coprime integers such that $t\neq0$ and $v_{p}\left( \dfrac{s}{t}\right) \leq0$. Then, $p\nmid s$.
Proof of Proposition 1. This is all well-known. Just in case, here are a few pointers:
Parts (a) and (b) of Proposition 1 are obvious in the case when one of $a$ and $b$ is $0$ (because $v_{p}\left( 0\right) =\infty\geq g$ for all $g\in\mathbb{Z}\cup\left\{ \infty\right\} $). Thus, they only need to be proven in the case when both $a$ and $b$ are nonzero. But in this case, they are exactly the parts (c) and (d) of Exercise 3.4.1 in my Introduction to Modern Algebra notes (version of 31 May 2019) (the numbering might change in the future, but you can always find the version of 31 May 2019 frozen on github). (Be warned that, in the latter notes, I use two different notations for what I am here calling $v_{p}\left( r\right) $: The first notation is "$v_{p}\left( r\right) $", which I use only in the case when $r$ is an integer; the second notation is "$w_{p}\left( r\right) $", which I use in the general case of rational $r$. The reason why I do this is to avoid a conflict of notations, even a theoretical one that doesn't actually happen; the notes are written for undergraduates.)
Proposition 1 (c) follows by induction on $k$. (The induction base uses $v_{p}\left( 1\right) =0$, while the induction step uses Proposition 1 (a).)
Proposition 1 (d) is Lemma 2.13.25 in my Introduction to Modern Algebra notes (version of 31 May 2019) (the numbering might change in the future, but you can always find the version of 31 May 2019 frozen on github). It is essentially a direct consequence of the definition of $v_{p}\left( n\right) $.
(e) Let $a,b\in\mathbb{Q}$ satisfy $v_{p}\left( a\right) >v_{p}\left( b\right) $. We must prove that $v_{p}\left( a+b\right) =v_{p}\left( b\right) $.
Assume the contrary. Thus, $v_{p}\left( a+b\right) \neq v_{p}\left( b\right) $. But Proposition 1 (b) yields $v_{p}\left( a+b\right) \geq\min\left\{ v_{p}\left( a\right) ,v_{p}\left( b\right) \right\} =v_{p}\left( b\right) $ (since $v_{p}\left( a\right) >v_{p}\left( b\right) $). Combined with $v_{p}\left( a+b\right) \neq v_{p}\left( b\right) $, this yields $v_{p}\left( a+b\right) >v_{p}\left( b\right) $. On the other hand, $v_{p}\left( \underbrace{-a}_{=a\left( -1\right) }\right) =v_{p}\left( a\left( -1\right) \right) =v_{p}\left( a\right) +v_{p}\left( -1\right) $ (by Proposition 1 (a), applied to $b=-1$). Since $-1$ is an integer, we have $v_{p}\left( -1\right) \geq0$. (Actually, $v_{p}\left( -1\right) =0$, but we don't need this.) Now, $v_{p}\left( -a\right) =v_{p}\left( a\right) +\underbrace{v_{p}\left( -1\right) }_{\geq0}\geq v_{p}\left( a\right) >v_{p}\left( b\right) $. Hence, both $v_{p}\left( a+b\right) $ and $v_{p}\left( -a\right) $ are $>v_{p}\left( b\right) $ (since we know that $v_{p}\left( a+b\right) >v_{p}\left( b\right) $). Thus, $\min\left\{ v_{p}\left( a+b\right) ,v_{p}\left( -a\right) \right\} >v_{p}\left( b\right) $ (since $\min\left\{ v_{p}\left( a+b\right) ,v_{p}\left( -a\right) \right\} $ must be one of the numbers $v_{p}\left( a+b\right) $ and $v_{p}\left( -a\right) $).
But $b=\left( a+b\right) +\left( -a\right) $. Hence, \begin{align*} v_{p}\left( b\right) =v_{p}\left( \left( a+b\right) +\left( -a\right) \right) \geq\min\left\{ v_{p}\left( a+b\right) ,v_{p}\left( -a\right) \right\} \end{align*} (by Proposition 1 (b), applied to $a+b$ and $-a$ instead of $a$ and $b$). This contradicts $\min\left\{ v_{p}\left( a+b\right) ,v_{p}\left( -a\right) \right\} >v_{p}\left( b\right) $. This contradiction shows that our assumption was wrong. Hence, Proposition 1 (e) is proven.
(f) Assume the contrary. Thus, $p\mid s$. Hence, $p^{1}=p\mid s$. But Proposition 1 (d) (applied to $i=1$ and $n=s$) shows that we have the equivalence $\left( p^{1}\mid s\right) \ \Longleftrightarrow\ \left( v_{p}\left( s\right) \geq1\right) $. Hence, we have $v_{p}\left( s\right) \geq1$ (since $p^{1}\mid s$).
But Proposition 1 (a) (applied to $a=\dfrac{s}{t}$ and $b=t$) yields $v_{p}\left( \dfrac{s}{t}\cdot t\right) =\underbrace{v_{p}\left( \dfrac {s}{t}\right) }_{\leq0}+v_{p}\left( t\right) \leq v_{p}\left( t\right) $. Hence, $v_{p}\left( t\right) \geq v_{p}\left( \underbrace{\dfrac{s}{t}\cdot t}_{=s}\right) =v_{p}\left( s\right) \geq1$. But Proposition 1 (d) (applied to $i=1$ and $n=t$) shows that we have the equivalence $\left( p^{1}\mid t\right) \ \Longleftrightarrow\ \left( v_{p}\left( t\right) \geq1\right) $. Hence, we have $p^{1}\mid t$ (since $v_{p}\left( t\right) \geq1$). Thus, $p=p^{1}\mid t$.
Now, $p$ is a common divisor of $s$ and $t$ (since $p\mid s$ and $p\mid t$). This shows that $s$ and $t$ have a common divisor larger than $1$ (since $p>1$); but this contradicts the fact that $s$ and $t$ are coprime. This contradiction shows that our assumption was false. Thus, Proposition 1 (f) is proven. $\blacksquare$
Next, we need a binomial identity:
Proposition 2. Let $n\in\mathbb{N}$. Then, \begin{align*} \sum\limits_{k=0}^{n}\dfrac{1}{\dbinom{n}{k}}=\dfrac{n+1}{2^{n+1}}\sum\limits_{k=1} ^{n+1}\dfrac{2^{k}}{k}. \end{align*}
Proposition 2 is the equality (7) in https://math.stackexchange.com/a/481686/ ; it is also Exercise 3.20 (b) in my Notes on the combinatorial fundamentals of algebra, version of 2019-01-10. $\blacksquare$
Next, we need two specific $3$-adic valuations:
Lemma 3. Let $k$ be a positive integer.
(a) We have $v_{3}\left( \dfrac{2^{k}}{k}\right) =-v_{3}\left( k\right) $.
(b) If $k$ is odd, then $v_{3}\left( \dfrac{2^{k}}{k}+\dfrac{2^{2k}} {2k}\right) =-v_{3}\left( k\right) $.
Proof of Lemma 3. Clearly, $2^{k}$ is an integer; thus, $v_{3}\left( 2^{k}\right) \in\mathbb{N}$. Similarly, $v_{3}\left( 1+2^{k-1}\right) \in\mathbb{N}$.
We don't have $3^{1}\mid2^{k}$ (since the only prime divisor of $2^{k}$ is $2$). Proposition 1 (d) (applied to $p=3$, $i=1$ and $n=2^{k}$) yields the equivalence $\left( 3^{1}\mid2^{k}\right) \ \Longleftrightarrow\ \left( v_{3}\left( 2^{k}\right) \geq1\right) $. Hence, we don't have $v_{3}\left( 2^{k}\right) \geq1$ (since we don't have $3^{1}\mid2^{k}$). In other words, we have $v_{3}\left( 2^{k}\right) <1$; thus, $v_{3}\left( 2^{k}\right) =0$ (since $v_{3}\left( 2^{k}\right) \in\mathbb{N}$).
(a) Proposition 1 (a) (applied to $p=3$, $a=\dfrac{2^{k}}{k}$ and $b=k$) yields $v_{3}\left( \dfrac{2^{k}}{k}\cdot k\right) =v_{3}\left( \dfrac{2^{k}}{k}\right) +v_{3}\left( k\right) $. Comparing this with $v_{3}\left( \underbrace{\dfrac{2^{k}}{k}\cdot k}_{=2^{k}}\right) =v_{3}\left( 2^{k}\right) =0$, we obtain $v_{3}\left( \dfrac{2^{k}} {k}\right) +v_{3}\left( k\right) =0$. Hence, $v_{3}\left( \dfrac{2^{k}} {k}\right) =-v_{3}\left( k\right) $. This proves Lemma 3 (a).
(b) Assume that $k$ is odd. Thus, $k-1$ is even. Now, $2\equiv -1\operatorname{mod}3$ and thus $2^{k-1}\equiv\left( -1\right) ^{k-1}=1\operatorname{mod}3$ (since $k-1$ is even). Hence, $1+\underbrace{2^{k-1}}_{\equiv1\operatorname{mod}3}\equiv1+1=2\not \equiv 0\operatorname{mod}3$, so that $3\nmid1+2^{k-1}$. But Proposition 1 (d) (applied to $p=3$, $i=1$ and $n=1+2^{k-1}$) yields the equivalence $\left( 3^{1}\mid1+2^{k-1}\right) \ \Longleftrightarrow\ \left( v_{3}\left( 1+2^{k-1}\right) \geq1\right) $. Hence, we don't have $v_{3}\left( 1+2^{k-1}\right) \geq1$ (since we don't have $3^{1}\mid1+2^{k-1}$ (because $3^{1}=3\nmid1+2^{k-1}$)). In other words, we have $v_{3}\left( 1+2^{k-1}\right) <1$; thus, $v_{3}\left( 1+2^{k-1}\right) =0$ (since $v_{3}\left( 1+2^{k-1}\right) \in\mathbb{N}$).
Now, Proposition 1 (a) (applied to $p=3$, $a=\dfrac{2^{k}}{k}$ and $b=1+2^{k-1}$) yields \begin{align*} v_{3}\left( \dfrac{2^{k}}{k}\left( 1+2^{k-1}\right) \right) =v_{3}\left( \dfrac{2^{k}}{k}\right) +\underbrace{v_{3}\left( 1+2^{k-1}\right) } _{=0}=v_{3}\left( \dfrac{2^{k}}{k}\right) =-v_{3}\left( k\right) \end{align*} (by Lemma 3 (a)). In view of \begin{align*} \dfrac{2^{k}}{k}\left( 1+2^{k-1}\right) =\dfrac{2^{k}}{k}+\underbrace{\dfrac {2^{k}}{k}\cdot2^{k-1}}_{=\dfrac{2^{2k-1}}{k}=\dfrac{2^{2k}}{2k}}=\dfrac {2^{k}}{k}+\dfrac{2^{2k}}{2k}, \end{align*} this rewrites as $v_{3}\left( \dfrac{2^{k}}{k}+\dfrac{2^{2k}}{2k}\right) =-v_{3}\left( k\right) $. This proves Lemma 3 (b). $\blacksquare$
Lemma 4. Let $n$ be a positive integer. Let $m$ be the largest nonnegative integer such that $n\geq3^{m}$. Set \begin{align*} a=\sum\limits_{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\3^{m}\nmid k} }\dfrac{2^{k}}{k}\qquad\text{and}\qquad b=\sum\limits_{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\3^{m}\mid k}}\dfrac{2^{k}}{k}. \end{align*} Then:
(a) We have $v_{3}\left( a\right) >-m$.
(b) We have $v_{3}\left( b\right) =-m$.
Proof of Lemma 4. (a) Define a subset $G$ of $\mathbb{Q}$ by \begin{align*} G=\left\{ r\in\mathbb{Q}\ \mid\ v_{3}\left( r\right) >-m\right\} . \end{align*} Then, $0\in G$ (since $v_{3}\left( 0\right) =\infty>-m$).
Next, we shall show that the set $G$ is closed under addition. Indeed, let $s,t\in G$. Then, $s\in G=\left\{ r\in\mathbb{Q}\ \mid\ v_{3}\left( r\right) >-m\right\} $; in other words, $s\in\mathbb{Q}$ and $v_{3}\left( s\right) >-m$. Likewise, $t\in\mathbb{Q}$ and $v_{3}\left( t\right) >-m$. Both numbers $v_{3}\left( s\right) $ and $v_{3}\left( t\right) $ are $>-m$ (since $v_{3}\left( s\right) >-m$ and $v_{3}\left( t\right) >-m$). Thus, $\min\left\{ v_{3}\left( s\right) ,v_{3}\left( t\right) \right\} >-m$ (since $\min\left\{ v_{3}\left( s\right) ,v_{3}\left( t\right) \right\} $ is one of these two numbers $v_{3}\left( s\right) $ and $v_{3}\left( t\right) $). Now, Proposition 1 (b) (applied to $3$, $s$ and $t$ instead of $p$, $a$ and $b$) yields $v_{3}\left( s+t\right) \geq\min\left\{ v_{3}\left( s\right) ,v_{3}\left( t\right) \right\} >-m$. In other words, $s+t\in G$ (by the definition of $G$). Now, forget that we fixed $s,t$. We thus have proven that $s+t\in G$ for all $s,t\in G$. In other words, the set $G$ is closed under addition. Hence, $G$ is a submonoid of the additive monoid $\left( \mathbb{Q},+\right) $ (since $0\in G$). Thus, any finite sum of elements of $G$ is an element of $G$.
Now, let $k\in\left\{ 1,2,\ldots,n\right\} $ be such that $3^{m}\nmid k$. Proposition 1 (d) (applied to $3$, $m$ and $k$ instead of $p$, $i$ and $n$) shows that we have the equivalence $\left( 3^{m}\mid k\right) \ \Longleftrightarrow\ \left( v_{3}\left( k\right) \geq m\right) $. Thus, we do not have $v_{3}\left( k\right) \geq m$ (since we do not have $3^{m}\mid k$ (because $3^{m}\nmid k$)). In other words, we have $v_{3}\left( k\right) <m$. But Lemma 3 (a) yields $v_{3}\left( \dfrac{2^{k}} {k}\right) =-\underbrace{v_{3}\left( k\right) }_{<m}>-m$. In other words, $\dfrac{2^{k}}{k}\in G$ (by the definition of $G$).
Forget that we fixed $k$. We thus have shown that $\dfrac{2^{k}}{k}\in G$ for each $k\in\left\{ 1,2,\ldots,n\right\} $ satisfying $3^{m}\nmid k$. In other words, all addends in the sum $\sum\limits_{\substack{k\in\left\{ 1,2,\ldots ,n\right\} ;\\3^{m}\nmid k}}\dfrac{2^{k}}{k}$ belong to $G$. Hence, $\sum\limits_{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\3^{m}\nmid k} }\dfrac{2^{k}}{k}$ is a finite sum of elements of $G$, and thus must be an element of $G$ itself (since any finite sum of elements of $G$ is an element of $G$). In other words, $\sum\limits_{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\3^{m}\nmid k}}\dfrac{2^{k}}{k}\in G$. Hence, $a=\sum\limits_{\substack{k\in \left\{ 1,2,\ldots,n\right\} ;\\3^{m}\nmid k}}\dfrac{2^{k}}{k}\in G$. In other words, $v_{3}\left( a\right) >-m$ (by the definition of $G$). This proves Lemma 4 (a).
(b) We are in one of the following two cases:
Case 1: We have $2\cdot3^{m}\leq n$.
Case 2: We have $2\cdot3^{m}>n$.
Let us first consider Case 1. In this case, we have $2\cdot3^{m}\leq n$. We have defined $m$ to be the largest nonnegative integer such that $n\geq3^{m}$. Thus, $n\geq3^{m}$ but $n<3^{m+1}$. Hence, $3^{m}\leq n$ and $3^{m+1}>n$. Thus, $3^{m}\in\left\{ 1,2,\ldots,n\right\} $ (since $3^{m}\leq n$) and $2\cdot3^{m}\in\left\{ 1,2,\ldots,n\right\} $ (since $2\cdot3^{m}\leq n$), but $3\cdot3^{m}\notin\left\{ 1,2,\ldots,n\right\} $ (since $3\cdot 3^{m}=3^{m+1}>n$). Hence, the only multiples of $3^{m}$ that belong to the set $\left\{ 1,2,\ldots,n\right\} $ are $3^{m}$ and $2\cdot3^{m}$. In other words, the only $k\in\left\{ 1,2,\ldots,n\right\} $ that satisfy $3^{m}\mid k$ are $3^{m}$ and $2\cdot3^{m}$.
But \begin{align*} b=\sum\limits_{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\3^{m}\mid k} }\dfrac{2^{k}}{k}=\dfrac{2^{3^{m}}}{3^{m}}+\dfrac{2^{2\cdot3^{m}}}{2\cdot 3^{m}} \end{align*} (since the only $k\in\left\{ 1,2,\ldots,n\right\} $ that satisfy $3^{m}\mid k$ are $3^{m}$ and $2\cdot3^{m}$). Hence, \begin{align*} v_{3}\left( b\right) =v_{3}\left( \dfrac{2^{3^{m}}}{3^{m}}+\dfrac {2^{2\cdot3^{m}}}{2\cdot3^{m}}\right) =-v_{3}\left( 3^{m}\right) \end{align*} (by Lemma 3 (b), applied to $k=3^{m}$), since $3^{m}$ is odd. Thus, \begin{align*} v_{3}\left( b\right) =-\underbrace{v_{3}\left( 3^{m}\right) }_{=m}=-m. \end{align*} Hence, Lemma 4 (b) is proven in Case 1.
Let us next consider Case 2. In this case, we have $2\cdot3^{m}>n$. Thus, $3^{m}\in\left\{ 1,2,\ldots,n\right\} $ (since $3^{m}\leq n$), but $2\cdot3^{m}\notin\left\{ 1,2,\ldots,n\right\} $ (since $2\cdot3^{m}>n$). Hence, the only multiple of $3^{m}$ that belongs to the set $\left\{ 1,2,\ldots,n\right\} $ is $3^{m}$. In other words, the only $k\in\left\{ 1,2,\ldots,n\right\} $ that satisfies $3^{m}\mid k$ is $3^{m}$.
But \begin{align*} b=\sum\limits_{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\3^{m}\mid k} }\dfrac{2^{k}}{k}=\dfrac{2^{3^{m}}}{3^{m}} \end{align*} (since the only $k\in\left\{ 1,2,\ldots,n\right\} $ that satisfies $3^{m}\mid k$ is $3^{m}$). Hence, \begin{align*} v_{3}\left( b\right) =v_{3}\left( \dfrac{2^{3^{m}}}{3^{m}}\right) =-v_{3}\left( 3^{m}\right) \end{align*} (by Lemma 3 (a), applied to $k=3^{m}$). Thus, \begin{align*} v_{3}\left( b\right) =-\underbrace{v_{3}\left( 3^{m}\right) }_{=m}=-m. \end{align*} Hence, Lemma 4 (b) is proven in Case 2.
We have now proven Lemma 4 (b) in both Cases 1 and 2. Hence, Lemma 4 (b) always holds. $\blacksquare$
Theorem 5. Let $n$ be a positive integer. Let $m$ be the largest nonnegative integer such that $n\geq3^{m}$. Then, \begin{align*} v_{3}\left( \sum\limits_{k=1}^{n}\dfrac{2^{k}}{k}\right) =-m. \end{align*}
Proof of Theorem 5. Define $a$ and $b$ as in Lemma 4. Then, Lemma 4 (a) yields $v_{3}\left( a\right) >-m$, but Lemma 4 (b) yields $v_{3}\left( b\right) =-m$. Thus, $v_{3}\left( a\right) >-m=v_{3}\left( b\right) $. Hence, Proposition 1 (e) (applied to $p=3$) yields $v_{3}\left( a+b\right) =v_{3}\left( b\right) =-m$.
But each $k\in\left\{ 1,2,\ldots,n\right\} $ satisfies either $3^{m}\nmid k$ or $3^{m}\mid k$ (but not both at the same time). Hence, we can split the sum $\sum\limits_{k=1}^{n}\dfrac{2^{k}}{k}$ as follows: \begin{align*} \sum\limits_{k=1}^{n}\dfrac{2^{k}}{k}=\underbrace{\sum\limits_{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\3^{m}\nmid k}}\dfrac{2^{k}}{k}} _{\substack{=a\\\text{(by the definition of }a\text{)}}}+\underbrace{\sum\limits _{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\3^{m}\mid k}}\dfrac{2^{k} }{k}}_{\substack{=b\\\text{(by the definition of }b\text{)}}}=a+b. \end{align*} Therefore, \begin{align*} v_{3}\left( \sum\limits_{k=1}^{n}\dfrac{2^{k}}{k}\right) =v_{3}\left( a+b\right) =-m. \end{align*} This proves Theorem 5. $\blacksquare$
Corollary 6. Let $n\in\mathbb{N}$. Then:
(a) We have \begin{align*} v_{3}\left( \sum\limits_{k=0}^{n}\dfrac{1}{\dbinom{n}{k}}\right) \leq0. \end{align*}
(b) Assume that $\sum\limits_{k=0}^{n}\dfrac{1}{\dbinom{n}{k}}=\dfrac{s}{t}$ for two coprime integers $s$ and $t$ satisfying $t\neq0$. Then, $3\nmid s$.
Proof of Corollary 6. (a) Let $m$ be the largest nonnegative integer such that $n+1\geq3^{m}$. Then, $n+1\geq3^{m}$ but $n+1<3^{m+1}$.
If we had $3^{m+1}\mid n+1$, then we would have $3^{m+1}\leq n+1$ (since $3^{m+1}$ and $n+1$ are positive integers), which would contradict $n+1<3^{m+1}$. Hence, we do not have $3^{m+1}\mid n+1$.
Theorem 1 (d) (applied to $3$, $m+1$ and $n+1$ instead of $p$, $i$ and $n$) yields the equivalence $\left( 3^{m+1}\mid n+1\right) \ \Longleftrightarrow\ \left( v_{3}\left( n+1\right) \geq m+1\right) $. Hence, we do not have $v_{3}\left( n+1\right) \geq m+1$ (since we do not have $3^{m+1}\mid n+1$). In other words, we have $v_{3}\left( n+1\right) <m+1$. Thus, $v_{3}\left( n+1\right) \leq m$ (since $v_{3}\left( n+1\right) $ and $m$ are integers).
Proposition 2 yields \begin{align*} \sum\limits_{k=0}^{n}\dfrac{1}{\dbinom{n}{k}}=\dfrac{n+1}{2^{n+1}}\sum\limits_{k=1} ^{n+1}\dfrac{2^{k}}{k}. \end{align*} Multiplying both sides of this equality by $\dfrac{2^{n+1}}{n+1}$, we obtain \begin{align*} \dfrac{2^{n+1}}{n+1}\sum\limits_{k=0}^{n}\dfrac{1}{\dbinom{n}{k}}=\dfrac{2^{n+1} }{n+1}\cdot\dfrac{n+1}{2^{n+1}}\sum\limits_{k=1}^{n+1}\dfrac{2^{k}}{k}=\sum\limits _{k=1}^{n+1}\dfrac{2^{k}}{k}. \end{align*} Thus, \begin{align*} v_{3}\left( \dfrac{2^{n+1}}{n+1}\sum\limits_{k=0}^{n}\dfrac{1}{\dbinom{n}{k} }\right) =v_{3}\left( \sum\limits_{k=1}^{n+1}\dfrac{2^{k}}{k}\right) =-m \end{align*} (by Theorem 5, applied to $n+1$ instead of $n$).
But Lemma 3 (a) (applied to $k=n+1$) yields \begin{align*} v_{3}\left( \dfrac{2^{n+1}}{n+1}\right) =-\underbrace{v_{3}\left( n+1\right) }_{\leq m}\geq-m. \end{align*} Now, Proposition 1 (a) (applied to $p=3$, $a=\dfrac{2^{n+1}}{n+1}$ and $b=\sum\limits_{k=0}^{n}\dfrac{1}{\dbinom{n}{k}}$) yields \begin{align*} v_{3}\left( \dfrac{2^{n+1}}{n+1}\sum\limits_{k=0}^{n}\dfrac{1}{\dbinom{n}{k} }\right) & =\underbrace{v_{3}\left( \dfrac{2^{n+1}}{n+1}\right) } _{\geq-m}+v_{3}\left( \sum\limits_{k=0}^{n}\dfrac{1}{\dbinom{n}{k}}\right) \\ & \geq-m+v_{3}\left( \sum\limits_{k=0}^{n}\dfrac{1}{\dbinom{n}{k}}\right) . \end{align*} Hence, \begin{align*} -m+v_{3}\left( \sum\limits_{k=0}^{n}\dfrac{1}{\dbinom{n}{k}}\right) \leq v_{3}\left( \dfrac{2^{n+1}}{n+1}\sum\limits_{k=0}^{n}\dfrac{1}{\dbinom{n}{k} }\right) =-m. \end{align*} Adding $m$ to both sides of this inequality, we find \begin{align*} v_{3}\left( \sum\limits_{k=0}^{n}\dfrac{1}{\dbinom{n}{k}}\right) \leq0. \end{align*} This proves Corollary 6 (a).
(b) We have $\dfrac{s}{t}=\sum\limits_{k=0}^{n}\dfrac{1}{\dbinom{n}{k}}$ and thus $v_{3}\left( \dfrac{s}{t}\right) =v_{3}\left( \sum\limits_{k=0}^{n}\dfrac {1}{\dbinom{n}{k}}\right) \leq0$ (by Corollary 6 (a)). Hence, Proposition 1 (f) (applied to $p=3$) yields $3\nmid s$. This proves Corollary 6 (b). $\blacksquare$
