Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{n\to \infty }\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right)\dots \left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right)\right\}$ is equal to

Imagine standing before a vast, infinite product. It looks like a mountain of numbers, a sequence of terms multiplying into the abyss. The problem asks us to evaluate the limit of the product defined as:

At first glance, your brain might scream to find a pattern or multiply the terms. However, in the world of JEE Advanced, the most elegant solutions often come from stepping back and observing the behavior of the components rather than brute-forcing the calculation.

Let us isolate the general term, $a_k=2^{\frac{1}{2}}-2^{\frac{1}{2k+1}}$. As $k$ marches toward infinity, the exponent $\frac{1}{2k+1}$ shrinks, getting closer and closer to $0$.

Consequently, the term $2^{\frac{1}{2k+1}}$ approaches $2^0$, which is $1$. Therefore, as $k$ gets larger, each term $a_k$ approaches:

This is a crucial realization. Every single term in our infinite product is eventually hovering around $0.414$.

Many students fall into the trap of trying to find a closed form for the product, perhaps using logarithms or looking for telescoping properties. But here, the secret lies in the magnitude.

We have established that $a_k$ is an increasing sequence that approaches $\sqrt{2}-1$. This means that for every $k$, the following inequality holds:

Let $c=\sqrt{2}-1$. Since $c\approx 0.414$, we know that $c<1$. This is the "Aha!" moment: we are multiplying $n$ terms, and every single one of them is strictly less than a constant $c$, where $c<1$.

If $a_k<c$ for all $k$, then the product $P_n=a_1\cdot a_2\cdot \cdots \cdot a_n$ must be less than $c\cdot c\cdot \cdots \cdot c$ ($n$ times). In other words:

We also know that $P_n$ is positive because each $a_k$ is positive. Thus, we have trapped our product within the following bounds:

Now, apply the limit as $n\to \infty$. We know that for any constant $c$ where $|c|<1$, the limit ${\lim }_{n\to \infty }{c}^n=0$.

By the Squeeze Theorem, since $P_n$ is squeezed between $0$ and a sequence that vanishes, $P_n$ itself must vanish. The final limit is:

It is a beautiful, clean result. The complexity of the product dissolves into nothingness because the terms are small enough to pull the entire product down to zero. Keep this in your toolkit: whenever you see a product of infinite terms, check if they are bounded by a value less than $1$. If they are, you have already won.