Prove by mathematical induction that (2n)! / (2²ⁿ(n!)²) ≤ 1 / (3n + 1)¹/² for all positive integers

Q: Why can't we just use Stirling's approximation to prove this?

Stirling's approximation is a powerful tool for limits and asymptotic behavior, but for a formal proof by induction, we need exact algebraic manipulation. Induction requires us to show the step-by-step transition from $k$ to $k+1$ without relying on approximations. We need to be precise, not just 'close enough'.

Q: Why is it safe to square both sides of the inequality?

Great question! Squaring an inequality is dangerous if the signs are unknown. However, here, $k$ is a positive integer. Therefore, all terms like $(2k+1)$, $(2k+2)$, and the square roots are strictly positive. Since both sides are positive, squaring them is a monotonic operation that preserves the inequality. Always check the signs before you square!

Q: I got stuck at the polynomial expansion. Is there a shortcut?

Algebraic stamina is part of the JEE game! While there are no 'magic' shortcuts, notice that both sides share the term $12k^3 + 28k^2 + 4$. Recognizing these common terms early saves you from tedious arithmetic and helps you focus on the core difference: $19k$ versus $20k$. Keep your eyes peeled for common factors!

Question

Prove by mathematical induction that $\frac{(2n)!}{2^{2n}(n!)^2} \le \frac{1}{(3n + 1)^{1/2}}$ for all positive integers $n$.

Sigma Percentile · Accepted Answer

### The Beauty of the Inductive Leap

Welcome, future engineer. Today, we are not just solving an inequality; we are embarking on a journey of logical certainty.

Mathematical induction is often misunderstood as a mechanical process, but it is actually a profound philosophical statement: if we can build a bridge from one step to the next, we can traverse the infinite. Let us prove that the following inequality holds for all positive integers $n$:

$$ \frac{(2n)!}{2^{2n}(n!)^2} \le \frac{1}{\sqrt{3n+1}} $$

### Phase 1: The Anchor (The Base Case)

Every great structure needs a foundation. We start with $n=1$.

On the left-hand side, we have:

$$ \frac{(2 	imes 1)!}{2^{2 	imes 1}(1!)^2} = \frac{2}{4} = \frac{1}{2} $$

On the right-hand side, we have:

$$ \frac{1}{\sqrt{3(1)+1}} = \frac{1}{\sqrt{4}} = \frac{1}{2} $$

Since $\frac{1}{2} \le \frac{1}{2}$, our base case is rock solid. We have established our starting point.

### Phase 2: The Leap of Faith (The Inductive Hypothesis)

Now, we assume the proposition $P(k)$ is true for some arbitrary positive integer $k$. This is our **Inductive Hypothesis**.

We accept that:

$$ \frac{(2k)!}{2^{2k}(k!)^2} \le \frac{1}{\sqrt{3k+1}} $$

This is our weapon. We do not know if it is true for all $k$, but we assume it is, and we use it to build the next step. This is the bridge we mentioned earlier.

### Phase 3: The Algebraic Dance

This is where the magic happens. We want to prove $P(k+1)$. Let us look at the left-hand side for $k+1$:

$$ \frac{(2k+2)!}{2^{2k+2}((k+1)!)^2} $$

We need to force this expression to reveal the $P(k)$ term we just assumed. By expanding the factorials, we write $(2k+2)!$ as $(2k+2)(2k+1)(2k)!$ and $((k+1)!)^2$ as $(k+1)^2(k!)^2$.

When we rearrange this, we get:

$$ \left[ \frac{(2k)!}{2^{2k}(k!)^2} ight] \cdot \frac{(2k+2)(2k+1)}{4(k+1)^2} $$

Look closely at that second fraction. It simplifies beautifully! By factoring out a $2$ from $(2k+2)$, we get:

$$ \frac{2(k+1)(2k+1)}{4(k+1)^2} = \frac{2k+1}{2k+2} $$

Now, our expression is simply the $P(k)$ term multiplied by $\frac{2k+1}{2k+2}$.

### Phase 4: The Final Victory

Using our hypothesis, we replace the $P(k)$ term with $\frac{1}{\sqrt{3k+1}}$. Now we need to show that:

$$ \frac{1}{\sqrt{3k+1}} \cdot \frac{2k+1}{2k+2} \le \frac{1}{\sqrt{3k+4}} $$

To prove this, we square both sides to eliminate the radicals, leading us to:

$$ (2k+1)^2(3k+4) \le (2k+2)^2(3k+1) $$

Expanding these polynomials is a test of patience, but look at the result:

$$ 12k^3 + 28k^2 + 19k + 4 \le 12k^3 + 28k^2 + 20k + 4 $$

The cubic and quadratic terms vanish! We are left with $19k \le 20k$, which is undeniably true for all $k \ge 1$.

We have done it. The bridge is complete, and the proof stands firm. You have mastered the logic of the infinite.

Animated Solution for Mathematics - Basic Mathematics: Prove by mathematical induction that $\frac{( 2 n )!}{2 ^{2 n} ( n ! ) ^{2}} \leq \frac{1}{( 3 n + 1 ) ^{1/2}}$ for all positive integers $n$ .

Visualized Solution

Defining the Proposition $P (n)$

The Base Case: $n = 1$

The Inductive Hypothesis

Setting up $P (k + 1)$

Algebraic Decomposition

Simplifying the Extra Terms

Applying the Inductive Hypothesis

Defining the Final Sub-problem

Squaring Both Sides

Cross-Multiplication

Polynomial Expansion (L.H.S.)

Polynomial Expansion (R.H.S.)

Final Comparison and Conclusion

The Sigma Insight: Linear Inequalities

The Beauty of the Inductive Leap

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Find the set of all $x$ for which $\frac{2 x}{2 x ^{2} + 5 x + 2} > \frac{1}{x + 1}$

If $S$ is the set of all real $x$ such that $\frac{2 x - 1}{2 x ^{3} + 3 x ^{2} + x}$ is positive, then $S$ contains

If $a, b, c$ are distinct +ve real numbers and $a^{2} + b^{2} + c^{2} = 1$ then $ab + b c + c a$ is

Let $y = \frac{( x + 1 ) ( x - 3 )}{( x - 2 )}$ . Find all the real values of $x$ for which $y$ takes real values.

Which of the following is the negation of the statement "for all $M > 0$ , there exists $x \in S$ such that $x \geq M$ "?

If $p \to (p \land \sim q)$ is false. Then the truth values of $p$ and $q$ are respectively

If $p \to (p \land \sim q)$ false, then the truth values of $p$ and $q$ are respectively :

The Boolean expression $(p \land \sim q) \Rightarrow (q \lor \sim p)$ is equivalent to:

Contrapositive of the statement "If two numbers are not equal, then their squares are not equal." Is:

Let $p$ : Ramesh listens to music. $q$ : Ramesh is out of his village $r$ : It is Sunday $s$ : It is Saturday Then the statement "Ramesh listens to music only if he is in his village and it is Sunday or Saturday" can be expressed as

Animated Solution for Mathematics - Basic Mathematics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Prove by mathematical induction that 22n(n!)2(2n)!​≤(3n+1)1/21​ for all positive integers n.