Animated Solution for Mathematics - Sequence and Series: Statement-1 : For every natural number $n\ge 2,\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}>\sqrt{n}$. Statement-2 : For every natural number $n\ge 2,\sqrt{n(n+1)}<n+1$.

A Journey into Monotonicity

Welcome, fellow traveler of the mathematical realm. Today, we are not just solving an inequality; we are peeling back the layers of a beautiful logical structure.

We have two statements, and our mission is to see how they dance together. Let us begin by dissecting the bedrock of this problem: Statement-2.

The Algebraic Bedrock

Statement-2 presents us with the inequality $\sqrt{n(n+1)}<n+1$. At first glance, it looks like a simple algebraic expression, but it holds the key to the entire problem.

Since we are given $n\ge 2$, we know that both sides are strictly positive. This is our green light to square both sides without fear of flipping the inequality sign.

When we square both sides, we get:

Now, look at the elegance of this simplification. We can divide both sides by $(n+1)$—which is perfectly safe because $n+1$ is positive—and we are left with $n<n+1$.

This is a universal truth! It is the most fundamental property of natural numbers. Thus, Statement-2 is undeniably true.

But more importantly, this inequality implies that $\sqrt{n}<\sqrt{n+1}$, which tells us that the square root function is strictly increasing. This is the 'soul' of the problem.

Visualizing the Sum

Now, let us turn our gaze to Statement-1: ${\sum }_{k=1}^n\frac{1}{\sqrt{k}}>\sqrt{n}$. This looks like a daunting sum, but let us transform it into a picture.

Imagine a coordinate plane. For each term $\frac{1}{\sqrt{k}}$, let us draw a rectangle with a width of $1$ and a height of $\frac{1}{\sqrt{k}}$. The total area of these rectangles is exactly our sum:

As $k$ increases, the height of our rectangles, $\frac{1}{\sqrt{k}}$, decreases. This is where the magic happens.

Because the square root function is strictly increasing (as we proved in Statement-2), its reciprocal, $\frac{1}{\sqrt{k}}$, must be strictly decreasing. This means every rectangle we draw is shorter than the one before it.

The very last rectangle, corresponding to $k=n$, is the shortest of them all, with a height of $\frac{1}{\sqrt{n}}$.

The Geometric Synthesis

Imagine drawing a horizontal dashed line at the height of this shortest rectangle, $y=\frac{1}{\sqrt{n}}$. Because every other rectangle is taller than this one, they all extend above this dashed line.

Now, construct a giant rectangle with this minimum height $\frac{1}{\sqrt{n}}$ that spans the entire width of our $n$ rectangles. The total width is $n$, and the height is $\frac{1}{\sqrt{n}}$.

The area of this 'bounding' rectangle is simply:

Since our original green rectangles are all taller than this bounding rectangle, their total area must be strictly greater than $\sqrt{n}$. And there it is! We have geometrically proven that ${\sum }_{k=1}^n\frac{1}{\sqrt{k}}>\sqrt{n}$.

The Logical Link

So, why is Statement-2 the correct explanation for Statement-1? Because the entire geometric argument relies on the fact that the sequence $\frac{1}{\sqrt{k}}$ is decreasing.

And how did we know it was decreasing? Because the square root function is increasing. And how did we prove that? Through the algebraic proof in Statement-2.

The two statements are not just related; they are two sides of the same coin. Both are true, and the second is the foundation upon which the first is built. Keep this perspective, and you will find that even the most complex JEE problems are just stories waiting to be told.