We actually need to prove by strong induction.
Suppose the result holds for all $2,3,4,...,n-1$.
Now let's look at the case of $n$.
We first look at the following sequences:
${a\over c}=1+{1\over3}+...+{1\over p}$ where $p$ is the largest odd number not exceeding $n$.
${b\over d}={1\over2}+...+{1\over q}$ where $q$ is the largest even number not exceeding $n$.
First we can conclude $b$ is odd and $d$ is even as
${b\over d}={1\over2}+...+{1\over q}={1\over2}(1+{1\over2}+{1\over3}...+{1\over({q\over2})})$.
By induction hypothesis we know for $q>2$, $(1+{1\over2}+{1\over3}...+{1\over({q\over2})})$ has odd numerator and even denominator and hence $d$ is even and $b$ is odd.
For the case where $q=2$, $b=1$ and $d=2$ so still $d$ is even and $b$ is odd.
Then let's look at $c$, we know $c$ must be odd because $1,3,5,...,p$ are all odd.
Now $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}={a\over c}+{b\over d}={ad+bc\over cd}$. The numerator is odd and the denominator is even and we are good.
No. You are assuming what you are trying to prove.
Consider this "proof" (the same as yours but I'm replacing the gray text with red text:
Theorem: The product of an even integer and an odd integer is $\color{gray}{\text{even}}$ $\color{red}{\text{odd}}$.
Proof: Let $a$
and $b$ be integers. Assume $a$ is even and $b$ is odd, so there exists an integer $k$ so that $a=2k$ and there exists an integer $q$ so that $b=2q+1$. If $a⋅b$ is $\color{gray}{\text{even}}$ $\color{red}{\text{odd}}$ then by definition of$\color{gray}{\text{even}}$ $\color{red}{\text{odd}}$ there exists an integer $r$ such that $\color{gray}{a*b=2r}$ $\color{red}{a*b=2r+1}$. So we have $a⋅b=(2p)(2q+1)=\color{gray}{2r}$ $\color{red}{2r+1}$, where $r$
is an integer.
Therefore,$ a⋅b$
is $\color{gray}{\text{even}}$ $\color{red}{\text{odd}}$.
Best Answer
You should try instead assuming the theorem is true for all $n \leq k$. (instead of just $n = k$).
Now, consider $k+1$. If $k+1$ is even, there is nothing to prove. If $k + 1$ is odd, we need to show $k + 2$ is even. Use your induction hypothesis here: in particular, you know that for $n = k-1$, if it is odd (can you show this?) then $n + 1 = k$ is even. What can you now conclude about $k + 2$?
Breaking this down into smaller steps.
Do you understand what the difference is in the induction hypothesis? (assuming for $n \leq k$ rather than $n = k$?)
Now we have to prove the theorem for the next larger number, $n = k + 1$. Either it is even or odd. What should we do if it is even?
If it is odd, we need to show the next bigger number, $n + 1 = k + 2$ is even.
At this point, what does the induction hypothesis say about $k -1$? (i.e. fill in the blanks: "if $k - 1$ is odd, then ____")
Is $k - 1$ odd? (otherwise the induction hypothesis doesn't say anything much).
What does this tell you about $k$? (in particular, is it even/odd?)
What does this tell you about $k + 2$? (the thing we're trying to prove something about).