Show that $H_n$ is almost always a non-terminating decimal. The following theorem and its proof is not so clear at some points:

Theorem.
$H_n$ is almost always a non-terminating decimal.
Proof.
We have that $H_1 = 1$, $H_2 = 1.5$ and $H_6 = 2.45$ and of course, since $H_n$ is always a fraction, its decimal expansion must either be finite, as with these examples, or infinitely recurring. The final surprise is that, apart from these three cases, all of the other $H_n$ are the infinitely recurring variety. Our proof of this remarkable fact will take us from comparatively shallow to very deep water, with the need of a most profound and significant result of number theory: the Bertrand Conjecture. In $1845$ the French mathematician Joseph Bertrand ($1822-1900$) conjectured that for every positive integer $n > 1$, there exists at least one prime $p$ satisfying $n < p < 2n$ (having verified it for $n < 3 000000$). He was not destined to provide a proof, but five years later the Russian mathematician Pafnuty Chebychev ($1821-1894$) was, and he was to come close to proving another of the great results of mathematics: the Prime Number Theorem, but more of that much later.
Firstly, it is clear that any number which can be represented as a finite decimal can be written as a fraction with denominator a power of $10$. Ten is two times five and so the denominator is a power of two times five, and, after possible cancellation, of the form $2^a5^b$. To show that $H_n $is not a finite decimal, it is enough to show that the denominator of $H_n$, when it is written as a single fraction, contains prime factors greater than $5$. Simply by writing out $H_3$, $H_4$ and $H_5$ we can establish that they are infinitely recurring; now write $H_n = a_n/b_n$, $n \ge 7$, where $a_n$ and $b_n$ are in their lowest terms. We need to show that $b_n$ is divisible by some prime $p \ge 7$. To that end we will prove that for all primes $p \in [\frac{1}{2} (n + 1), n]$, $p$ divides $b_n$ and do so by induction on $n$. For $n = 1$ the interval is $[4, 7]$, the set of primes ${\{5, 7}\}$ and since $H_7 = \frac{363}{140}$ we are done. Now assume the result for $n$, then we need to show that for all $p \in [\frac{1}{2} (n + 2), n + 1]$, $p$ divides $b_{n+1}$, where $$\dfrac{a_{n+1}}{b_{n+1}} = \dfrac{a_n}{b_n} + \dfrac{1}{n+1} = \dfrac{a_n(n+1)+b_n}{b_n(n+1)}.$$ Since this new interval can only add $n + 1$ to the list of primes and since if $n + 1$ is prime, $b_{n+1} = b_n(n + 1)$ is incapable of cancellation with the $a_{n+1}$, we have what we need and the result is true by induction. The Bertrand Conjecture guarantees that the set of intervals $[p, 2p-1]$ for $p \ge 7$ overlap and therefore contain every integer $n \ge 7$, since it guarantees a prime between every pair $p$ and $2p$. Now we have all that we need. If $n \ge 7$, there is a prime $p \ge 7$ such that $n \in [p, 2p-1]$, which means that $p \in [\frac{1}{2}(n + 1), n]$ and so divides $b_n$.

I can't fully understand the proof. Having answers to the following questions would help me through the proof though:
a. As text "if" $n+1$ is a prime, but it doesn't evaluate if $n+1$ is not a prime; then it is divisible by primes that some of them may coincide with the list of primes of decomposition of $a_n$ and $b_n$ which makes the problem more complicated.
b. If $n+1$ is a prime and if $n+1$ is not in the list of prime factors of $a_n$ and $b_n$ then I understand that $\frac{a_{n+1}}{b_{n+1}}$ is such that $a_{n+1}$ and $b_{n+1}$ in their lowest terms. But how to prove that non of the prime factors of $a_n$ and $b_n$ has $n+1$?
c. The induction process is not clear. What is the generalization 'recipe' for going from $k$ to $k+1$?
Thank you.
 A: The proof is not always very well written, so I will try to write it out in a way that will answer your questions.
Lemma Let $n\ge7$.  Define $H_n=\frac{a_n}{b_n}$, where $\gcd(a_n,b_n)=1$.  Suppose $p$ is a prime such that $\frac12(n+1)\le p\le n$.  Then $p$ divides $b_n$.  
Proof.  Induction on $n$.  For the $n=7$ case, note that $H_7=\frac{363}{140}$.  $\frac12(7+1)=4$ and the only primes $p$ such that $4\le p\le 7$ are $5$ and $7$, which both divide $140$.  
Now suppose that we know the lemma is true for $n=7,\dots,k$.  We prove that it is true for $n=k+1$.  
First note that
$$
\frac{a_{k+1}}{b_{k+1}}=\frac{a_k}{b_k}+\frac{1}{k+1}=\frac{a_k(k+1)+b_k}{b_k(k+1)}
$$
Now it is not necessarily true that $a_{k+1}=a_k(k+1)+b_k$ and that $b_{k+1}=b_k(k+1)$, since these values may have common factors that we can cancel out.  Nevertheless, we do know that $a_{k+1}$ must divide $a_k(k+1)+b_k$ and that $b_{k+1}$ must divide $b_k(k+1)$.
Now suppose that $p$ is a prime such that $\frac12((k+1)+1)\le p\le k+1$.  We have two cases:
Case 1: $p<k+1$.  In that case, we know that $p$ does not divide $k+1$ (since if it did then it would be at most $\frac12(k+1)$).  Now we know that $\frac12((k+1)+1)\le p\le k$; in particular, $\frac12(k+1)\le p\le k$, so by the induction hypothesis, $p$ divides $b_k$.  
Since $a_k,b_k$ are coprime, $p$ does not divide $a_k$.  It follows that $p$ does not divide $a_k(k+1)+b_k$, but does divide $b_k(k+1)$.  Therefore, we conclude that $p$ does not cancel out and therefore that $p$ divides $b_{k+1}$.  
Case 2: $p=k+1$.  We claim that $p$ does not divide $b_k$.  Indeed, this is easily seen:
$$
\frac{a_k}{b_k}=1+\frac12+\frac13+\cdots+\frac1k=\frac{\text{something}}{k!}
$$
and $p$ cannot divide $k!$.  Therefore, $p$ cannot divide $a_k(k+1)+b_k=a_kp+b_k$.  It follows that $p$ does not cancel out, so $p$ divides $b_{k+1}$.  $\Box$
Theorem Let $n\ge7$.  Then there is a prime $p\ge7$ such that $p$ divides $b_n$.  
Proof. Applying Bertrand's postulate to the number $\frac12(n+1)$ (or $\frac12n$ if $n$ is even), we deduce that there is a prime $p$ such that $\frac12(n+1)\le p\le n$.  Now use the preceding lemma.  $\Box$
