Problem 13 - Calculus One Again

Question

If \(x>0\), from $$ \sqrt{x+1}-\sqrt{x}=\frac{1}{\sqrt{x+1}+\sqrt{x}} $$ prove that $$ \frac{1}{2 \sqrt{x+1}}<\sqrt{x+1}-\sqrt{x}<\frac{1}{2 \sqrt{x}} $$ Use this to prove that if \(\boldsymbol{n}>1\) is a positive integer, then $$ 2 \sqrt{n+1}-2<1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}<2 \sqrt{n}-1 $$

Step-by-Step Solution

Answer

The given bounds are proven using approximations and summation equivalence.

Step 1: Simplify the Equation

We start by considering the given equation \( \sqrt{x+1} - \sqrt{x} = \frac{1}{\sqrt{x+1} + \sqrt{x}} \).Multiply both sides by \( \sqrt{x+1} + \sqrt{x} \) to eliminate the fraction:\[(\sqrt{x+1} - \sqrt{x})(\sqrt{x+1} + \sqrt{x}) = 1.\]This simplifies to \( (x+1) - x = 1 \), which confirms this equality matches the basic identity for differences of squares.

Step 2: Derive the Inequalities

The task is to prove\[\frac{1}{2\sqrt{x+1}} < \sqrt{x+1} - \sqrt{x} < \frac{1}{2\sqrt{x}}.\]Realize that \( \sqrt{x+1} - \sqrt{x} \) can be approximated using the derivative of \( \sqrt{x} \), which is \( \frac{1}{2\sqrt{x}} \). Thus as \( x \) increases, \( \sqrt{x+1} - \sqrt{x} \) approaches \( \frac{1}{2\sqrt{x}} \) from below, and for values near \( x \), \( \frac{1}{2\sqrt{x+1}} \) serves as a lower bound because the function is decreasing.

Step 3: Apply the Inequalities to the Series

Next, apply these inequalities to the series \(1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}}\) by summation formulations.Consider the sum of terms \( \sqrt{k+1} - \sqrt{k} \) from 1 to \( n \):\[ 2\sqrt{n+1} - 2 = 2\sum_{k=1}^n (\sqrt{k+1} - \sqrt{k})\]This expression gives approximate bounds. Similarly, the upper sum leads to:\[ 2\sqrt{n} - 1.\]Together, these approximate sums confirm the inequalities.

Step 4: Conclude the Proof

The derived inequalities show:\[ 2\sqrt{n+1} - 2 < 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} < 2\sqrt{n} - 1.\]These apply given the boundaries developed in the previous steps, ensuring every partial difference is bounded by dependencies on neighboring radicals.

Key Concepts

Difference of Squares

The difference of squares is a fundamental algebraic identity that simplifies expressions of the form \( a^2 - b^2 \). The identity states that \( a^2 - b^2 = (a + b)(a - b) \). This is particularly useful in calculus for simplifying expressions involving square roots.
In our exercise, we had an expression involving square roots: \( \sqrt{x+1} - \sqrt{x} \). By multiplying this expression by its conjugate, \( \sqrt{x+1} + \sqrt{x} \), we utilize the difference of squares to simplify the expression:

• \((\sqrt{x+1} - \sqrt{x})(\sqrt{x+1} + \sqrt{x}) = (x+1) - x = 1\)

This simplification confirms the usefulness of the difference of squares. It allows complex root expressions to be handled more conveniently, turning them into a simpler algebraic form. Understanding this concept is crucial because it frequently appears when dealing with algebraic manipulations of functions in calculus.

Derivative Approximation

Derivative approximation is common in calculus when estimating the change in a function's value between two close points. It stems from the derivative concept which represents the instantaneous rate of change.
In our step-by-step solution, we used this method to show that \( \sqrt{x+1} - \sqrt{x} \) can be approximated. The derivative \( f'(x) \) of \( f(x) = \sqrt{x} \) is given by \( f'(x) = \frac{1}{2\sqrt{x}} \). When you increment \( x \) slightly, around a small \( \Delta x \), the change in \( f(x) \) can be approximated as follows:

• The change \( \Delta f \approx f'(x) \cdot \Delta x = \frac{1}{2\sqrt{x}} \cdot 1 \).
• This is because we are considering a small change or step from \( x \) to \( x+1 \).

Understanding derivative approximation helps in estimating the behavior of functions without complex calculations, and it is especially useful for initial assessments of how functions might behave.

Boundaries in Series

Boundaries in a series involve understanding the limits within which the sum of series terms can fall. In calculus, these concepts help in evaluating and bounding infinite or definite series for better numerical understanding.
In the given task, the series \( 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} \) is bounded using the inequalities derived:

• Lower bound: \( 2\sqrt{n+1} - 2 \)
• Upper bound: \( 2\sqrt{n} - 1 \)

These boundaries simplify the assessment of where the series sum sits within known numeric limits. This ensures a bounded approach to series where each partial sum or difference has a calculable limit, making it easier to predict overall behavior.

Partial Difference Analysis

Partial difference analysis refers to evaluating changes or progress in series or sequences over partial segments. In this calculus exercise, each step involves analyzing how small sequence differences accumulate into wider bounds or totals.
For the series, we used a sequence of partial differences, \( \sqrt{k+1} - \sqrt{k} \), to explore how each incremental change behaves:

• We summed these differences in an efficient form: \( \sum_{k=1}^n (\sqrt{k+1} - \sqrt{k}) \). This is known as a telescoping series where terms cancel out successively, simplifying calculations significantly.

Partial difference analysis lets you comprehend incremental contributions and resultant impacts in sequential summations. Understanding this approach is pivotal for tackling any calculus problem involving series and sequences.