<s>[INST] Prove that for all $n$ there exist positive integers $x_1, ..., x_n$ such that $\frac{1}{x_1} + \frac{1}{2x_2} + ... + \frac{1}{nx_n} = 1$ Prove that for all $n$ there exist positive integers $x_1, ..., x_n$ such that $\frac{1}{x_1} + \frac{1}{2x_2} + ... + \frac{1}{nx_n} = 1$
I actually managed to write a short proof, but I'd appreciate it if someone could check it for me and offer some feedback. I'm not very used to induction so I'm not sure the proof is very clear or correct. Hope that's appropriate for this forum!
My intuition was that starting with $n = 2 \rightarrow\frac{1}{1\times2} + \frac{1}{2\times1} = 1$ you can always add one more term by making $x_n = n + 1$ and $x_{n+1} = 1$, such as in
$\frac{1}{1\times2} + \frac{1}{2\times1} = \frac{1}{1\times2} + \frac{1}{2\times3} + \frac{1}{3\times1} = \frac{1}{1\times2} + \frac{1}{2\times3} + \frac{1}{3\times4} + \frac{1}{4\times1}$ = ... = 1
So here's the proof:
Statement: for all $n$ there exist positive integers $x_1, ..., x_n$ such that $\frac{1}{x_1} + \frac{1}{2x_2} + ... + \frac{1}{nx_n} = 1$
Proof (by induction):
   Base case:
     For $n = 1$, we have that $\frac{1}{x_1} = 1$ for $x_1 = 1$
   Inductive step:
     If for some $n$:
     $\frac{1}{x_1} + \frac{1}{2x_2} + … + \frac{1}{nx_n} = 1$, for certain $x_1,x_2,...,x_n$
     and we also have that $x_n = 1$
     It follows that:
     $\frac{1}{x_1} + \frac{1}{2x_2} + … + \frac{1}{n} = 1$
     $\frac{1}{x_1} + \frac{1}{2x_2} + … + \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+1} = 1$
     $\frac{1}{x_1} + \frac{1}{2x_2} + … + \frac{n + 1 - n}{n(n+1)} + \frac{1}{n+1} = 1$
     $\frac{1}{x_1} + \frac{1}{2x_2} + … + \frac{1}{n(n+1)} + \frac{1}{n+1} = 1$
     Therefore, if $y_i = x_i$ for $1 \leq i \leq n-1$, $y_n = n + 1$ and $y_{n+1} = 1$, it follows that:
     $\frac{1}{y_1} + \frac{1}{2y_2} + … + \frac{1}{ny_n} + \frac{1}{(n+1)y_{n+1}}= 1$
$$\tag*{$\blacksquare$}$$
Thanks in advance!
 [/INST] I think the idea is correct, but you do not need induction.
You suggested that $\frac{1}{2*1}+\frac{1}{3*2}+...+ \frac{1}{n(n-1)}+\frac{1}{n}=1$. This is true: the sum is $\frac11-\frac12+\frac12-\frac13+....+\frac{1}{n-1}-\frac{1}{n}+\frac{1}{n}=1$
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