Induction for Products
The product $\prod_{r=1}^{n}\!\left(1+\tfrac{1}{r}\right)$ multiplies together fractions that look complicated, yet the result collapses to the elegant $n+1$. Telescoping products and factorial identities are the two main families here. The inductive step always boils down to one move: peel off the $(k+1)$th factor, substitute the IH, and simplify.
Compute $\left(1+\dfrac{1}{1}\right)\!\left(1+\dfrac{1}{2}\right)\!\left(1+\dfrac{1}{3}\right)\!\left(1+\dfrac{1}{4}\right)$ step by step. Before reading on — what pattern do you notice in your answer?
Every product induction proof uses two moves: split off the $(k+1)$th factor from the product, then substitute the inductive hypothesis for the remaining $k$-factor product.
The fundamental splitting identity for products:
$\prod_{r=1}^{k+1} a_r = \left(\prod_{r=1}^{k} a_r\right) \cdot a_{k+1}$
Apply IH to the first factor, then simplify the product of the IH result with $a_{k+1}$. Telescoping products often cancel dramatically — write each $a_r$ as a fraction to spot this.
Key facts
- $\prod_{r=1}^{n} a_r = a_1 \cdot a_2 \cdots a_n$ (product notation)
- $\prod_{r=1}^{k+1} a_r = \left(\prod_{r=1}^{k} a_r\right) \cdot a_{k+1}$ (splitting identity)
- $(k+1) \cdot k! = (k+1)!$ (factorial recurrence)
Concepts
- Why the inductive step for products requires splitting off the last factor
- How telescoping products collapse when written as fractions
- The connection between products of consecutive integers and factorials
Skills
- Prove a telescoping product formula by induction
- Prove a factorial product identity by induction
- Identify and fix errors in incomplete product induction proofs
The product $\prod_{r=1}^{n} a_r$ means $a_1 \cdot a_2 \cdots a_n$. To prove product formulas by induction, the inductive step uses the splitting identity:
You then substitute the inductive hypothesis for $\prod_{r=1}^{k} a_r$ and simplify with the explicit $(k+1)$th factor.
Two important families:
- Telescoping: each $a_r = \frac{f(r+1)}{f(r)}$, so the product collapses. Example: $\prod_{r=1}^{n}\!\left(1+\frac{1}{r}\right) = \frac{2}{1}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdots\frac{n+1}{n} = n+1$.
- Factorial: products of consecutive integers. Example: $\prod_{r=1}^{n}(2r) = 2 \cdot 4 \cdot 6 \cdots 2n = 2^n \cdot n!$.
The product $\prod_{r=1}^{n} a_r$ means $a_1 \cdot a_2 \cdots a_n$. To prove product formulas by induction, the inductive step uses the splitting identity :
Pause — copy the product notation $\prod_{r=1}^n a_r$ and the splitting identity $\prod_{r=1}^{k+1}=\prod_{r=1}^k\cdot a_{k+1}$ into your book.
Quick check: In the inductive step of a product induction proof, what is the correct first line?
We just saw that $\prod_{r=1}^n a_r=a_1\cdot a_2\cdots a_n$ and the inductive step uses $\prod_{r=1}^{k+1}=\prod_{r=1}^k\cdot a_{k+1}$. That raises a question: applying this to $\prod_{r=1}^{k+1}(1+1/r)$ and substituting the hypothesis $k+1$, how does multiplying by $\frac{k+2}{k+1}$ give exactly $k+2$? This card answers it → $(k+1)\cdot\frac{k+2}{k+1}=k+2$ by cancellation.
Prove that $\displaystyle\prod_{r=1}^{n}\!\left(1+\dfrac{1}{r}\right) = n+1$ for all positive integers $n$.
LHS $= 1 + \dfrac{1}{1} = 2$. RHS $= 1 + 1 = 2$. LHS = RHS. ✓
Assume $\displaystyle\prod_{r=1}^{k}\!\left(1+\dfrac{1}{r}\right) = k+1$ for some $k \geq 1$.
$\displaystyle\prod_{r=1}^{k+1}\!\left(1+\dfrac{1}{r}\right) = \left[\prod_{r=1}^{k}\!\left(1+\dfrac{1}{r}\right)\right]\cdot\left(1+\dfrac{1}{k+1}\right)$ (splitting identity)
$= (k+1) \cdot \dfrac{k+2}{k+1}$ (substitute IH; write the new factor as a single fraction)
$= k+2$ (cancel $(k+1)$)
This equals $(k+1)+1$, the formula with $n=k+1$. ✓
By the principle of mathematical induction, the result holds for all $n \geq 1$. ∎
Prove that $\displaystyle\prod_{r=1}^{n}\!\left(1+\dfrac{1}{r}\right) = n+1$ for all positive integers $n$.
Pause — copy the telescoping product proof for $\prod_{r=1}^n(1+1/r)=n+1$, highlighting the one-line inductive step $(k+1)\cdot\frac{k+2}{k+1}=k+2$ into your book.
Did you get this? True or false: $\prod_{r=1}^{n}\!\left(1+\dfrac{1}{r}\right) = n+1$ gives the value $5$ when $n = 4$.
Worked examples · 3 in a row, reveal as you go
Prove by induction that $\displaystyle\prod_{r=1}^{n}(2r) = 2^n \cdot n!$ for all positive integers $n$.
Prove by induction that $\displaystyle\prod_{r=2}^{n}\!\left(1 - \dfrac{1}{r^2}\right) = \dfrac{n+1}{2n}$ for all integers $n \geq 2$.
Prove by induction that $\displaystyle\prod_{r=1}^{n} \dfrac{r}{r+1} = \dfrac{1}{n+1}$ for all positive integers $n$.
Fill the gap: In proving $\prod_{r=1}^{n}(2r) = 2^n \cdot n!$, the inductive step gives $2^k \cdot k! \cdot 2(k+1) = 2^{k+1} \cdot$ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $\prod_{r=1}^{3}(2r) = 48$.
Activities · practice with the ideas
Compute $\prod_{r=1}^{4}\!\left(1+\frac{1}{r}\right)$ by expanding all four factors and verify it equals $5 = 4+1$.
Write out the full inductive step for $\prod_{r=1}^{n}\!\left(1+\frac{1}{r}\right) = n+1$. Start with "Split off the $(k+1)$th factor…"
Evaluate $\prod_{r=1}^{4}(2r)$ and verify it equals $2^4 \cdot 4!$.
Explain why the factor $1 - \frac{1}{r^2}$ should be rewritten as $\frac{(r-1)(r+1)}{r^2}$ before attempting the inductive step.
A student claims $\prod_{r=1}^{3}\frac{r}{r+1} = \frac{1}{3}$. Are they correct? What is the right answer?
Odd one out: Three of these product facts are correct. Which one is NOT?
Earlier you computed $\left(1+\frac{1}{1}\right)\!\left(1+\frac{1}{2}\right)\!\left(1+\frac{1}{3}\right)\!\left(1+\frac{1}{4}\right)$ and noticed a pattern.
The result is $\frac{2}{1}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4} = 5$. The formula says $n+1 = 4+1 = 5$ ✓. The reason is telescoping: every intermediate numerator cancels with the next denominator, leaving only the very first denominator ($1$) and the very last numerator ($n+1$). The induction proof is essentially just making this cancellation rigorous for all $n$.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Prove by induction that $\displaystyle\prod_{r=1}^{n}\!\left(1+\dfrac{1}{r}\right) = n+1$ for all positive integers $n$. (3 marks)
Q2. Prove by induction that $\displaystyle\prod_{r=2}^{n}\!\left(1-\dfrac{1}{r^2}\right) = \dfrac{n+1}{2n}$ for all integers $n \geq 2$. (3 marks)
Q3. (a) Prove by induction that $\displaystyle\prod_{r=1}^{n}(2r) = 2^n \cdot n!$ for all positive integers $n$. (3 marks)
(b) Hence evaluate $\displaystyle\prod_{r=1}^{5}(2r)$ without expanding each factor. (1 mark)
Comprehensive answers (click to reveal)
Activity answers: 1. $(2/1)(3/2)(4/3)(5/4) = 5$ ✓ · 2. Split, apply IH to get $(k+1)\cdot\frac{k+2}{k+1} = k+2$ · 3. $2\cdot4\cdot6\cdot8 = 384 = 16\cdot24 = 2^4\cdot4!$ ✓ · 4. Factored form allows cancellation of $(k+1)$ terms · 5. $\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4} = \frac{1}{4}$, not $\frac{1}{3}$; formula gives $\frac{1}{4}$.
Q1 (3 marks): Base: $n=1$: $2=2$ ✓ [1]. IH: $\prod_{r=1}^{k}(1+\frac{1}{r})=k+1$. Step: $(k+1)\cdot\frac{k+2}{k+1} = k+2$ [1]. Conclusion [1]. ∎
Q2 (3 marks): Base $n=2$: $\frac{3}{4}=\frac{3}{4}$ ✓ [1]. Step: $\frac{k+1}{2k}\cdot\frac{k(k+2)}{(k+1)^2}=\frac{k+2}{2(k+1)}$ [1]. Conclusion [1]. ∎
Q3 (4 marks): Base: $2=2^1\cdot1!$ ✓ [1]. Step: $2^k\cdot k!\cdot2(k+1) = 2^{k+1}(k+1)!$ [1]. Conclusion [1]. (b) $2^5\cdot5! = 32\cdot120 = \mathbf{3840}$ [1].
Five timed questions on product induction. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering product and induction questions. Lighter alternative to the boss.
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