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I'm having a hard time visualizing how to prove the following by induction:

For every positive integer $n$, let $[n]$ denote the set $\{1,\ldots,n\}$.

Let $A$ be a set. Use the notation $P(A)$ to indicate the power set of $A$, which consists of all subsets of $A$. For example, if $A=\{0,1\}$, then $P(A)=\big\{\{\},\{0\},\{1\},\{0,1\}\big\}$. Consider $Q(n) = P\big(\{1,\ldots,n\}\big)-\big\{\{\}\big\}$ and use an inductive argument to show that the sum

$$\sum_{S\in Q(n)}\frac1{\prod S}=n\;.$$

For example, the expansion for $n=3$ is

$$\frac11+\frac12+\frac13+\frac1{1\cdot2}+\frac1{1\cdot3}+\frac1{2\cdot3}+\frac1{1\cdot2\cdot3}=3\;.$$

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Tommy
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  • I modified the statement of the question to match the example; please check to see that it’s correct. In case the notation is unfamiliar, $\prod S$ is the product of the members of the set $S$. – Brian M. Scott Jan 10 '15 at 17:10
  • Yes that is correct. Thank you very much. I'm new here and haven't figured out how to inset equations yet. – Tommy Jan 10 '15 at 18:06
  • You’re very welcome. Here is a reference and tutorial. – Brian M. Scott Jan 10 '15 at 18:09

1 Answers1

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HINT: Suppose that it’s true for $n$:

$$\sum_{S\in Q(n)}\frac1{\prod S}=n\;.\tag{1}$$

We want to show that it’s true for $n+1$, i.e., that

$$\sum_{S\in Q(n+1)}\frac1{\prod S}=n+1\;.\tag{2}$$

Every non-empty subset of $[n]$ is a non-empty subset of $[n+1]$, so $Q(n)\subseteq Q(n+1)$. What are the other non-empty subsets of $[n+1]$? They’re exactly the subsets of $[n+1]$ that include the integer $n+1$. In other words, they’re the set $\{n+1\}$ and the sets of the form $S\cup\{n+1\}$ with $S\in Q(n)$. This means that

$$\sum_{S\in Q(n+1)}\frac1{\prod S}=\sum_{S\in Q(n)}\frac1{\prod S}+\sum_{S\in Q(n)}\frac1{(n+1)\prod S}+\frac1{n+1}\;;$$

why? Now use the induction hypothesis $(1)$ to show that $(2)$ holds.

Brian M. Scott
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  • Thanks for the hint.

    I understand the induction that

    $\sum_{S\in Q(n+1)}\frac1{\prod S}=\sum_{S\in Q(n)}\frac1{\prod S}+\sum_{S\in Q(n)}\frac1{(n+1)\prod S}+\frac1{n+1};$

    $\sum_{S\in Q(n+1)}\frac1{\prod S}=n+\sum_{S\in Q(n)}\frac1{(n+1)\prod S}+\frac1{n+1};$

    Although I'm still unclear how to show

    $\sum_{S\in Q(n)}\frac1{(n+1)\prod S}+\frac1{n+1};$ = 1

     – Tommy Jan 10 '15 at 19:53
  • @Tommy: Does it help if I point out that $$\sum_{S\in Q(n)}\frac1{(n+1)\prod S}=\frac1{n+1}\sum_{S\in Q(n)}\frac1{\prod S};?$$ – Brian M. Scott Jan 10 '15 at 19:54
  • Yes! That was what I was missing! Thank you very much! I've been stuck on this 2 days now. It has been 7 years since I last took a discreet mathematics class, needless to say I'm more than a little rusty. I greatly appreciate your help! – Tommy Jan 10 '15 at 20:14
  • @Tommy: You’re very welcome! – Brian M. Scott Jan 10 '15 at 20:17