Question

Prove that, for any \(n \in \mathbb{N}\), $$ 1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}>\log (n+1) $$

Short answer

Based on the provided step by step solution, answer the following question: **Question:** Prove that the sum of the reciprocals of the natural numbers up to n is greater than the logarithm of (n+1) for all natural numbers n. **Answer:** To prove this, we use mathematical induction. First, we verify the inequality for the base case when n = 1. We find that 1 > log(2), which confirms the inequality holds for the base case. Then, we assume the inequality holds for n = k and move to prove it for n = k + 1. By using the inductive hypothesis and constructing a lower bound for the logarithm function, we successfully show that the inequality holds for n = k + 1. Since the inequality holds for both the base case and the inductive step, we conclude that the sum of the reciprocals of the natural numbers up to n is greater than the logarithm of (n+1) for all natural numbers n.

Step-by-step solution

Base Case (n = 1)

First, we will verify the given inequality for the base case, when n=1. In this case, the sum has only one term, which is \(\frac{1}{1}=1\). According to the inequality, we need to check if this sum is greater than \(\log(1+1) = \log(2)\). It is true that \(1 > \log(2)\), as the natural logarithm of 2 is around 0.69. Therefore, the inequality is true for the base case.

Inductive Step (Assume inequality holds for n = k)

Now, let's suppose the inequality is true for n=k, meaning: $$ 1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{k}>\log(k+1) $$ We want to prove that the inequality also holds for n = k+1. In other words, we want to show that: $$ 1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{k}+\frac{1}{k+1}>\log((k+1)+1) = \log(k+2) $$

Use the Inductive Hypothesis

Using the induction hypothesis, we replace the inequality with the known inequality for n=k and add \(\frac{1}{k+1}\) to both sides of the inequality: $$ 1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{k}+\frac{1}{k+1}>\log(k+1)+\frac{1}{k+1} $$

Create a Lower Bound for the Logarithm

Now, we will define a lower bound for the \(\log\) of \(k+2\), which will help us prove the desired inequality. Since \(\log(x)\) is a concave function, we know that for any \(x\) in the interval from \(k+1\) to \(k+2\), \(\log(x)\) will be greater than or equal to the value of the tangent line of the function at the point \(x=k+1\). Let's find the equation for this tangent line by taking the derivative of \(\log(x)\) and finding the slope at the point \(x=k+1\). The derivative of \(\log(x)\) is \(\frac{1}{x}\), so the slope of the tangent line at the point \(x=k+1\) is \(\frac{1}{k+1}\). Thus, the equation of the tangent line is given by: $$ y = \frac{1}{k+1}(x-(k+1))+\log(k+1) $$ When \(x = k+2\), we get: $$ y = \frac{1}{k+1}((k+2) - (k+1)) + \log(k+1) = \frac{1}{k+1}+\log(k+1) $$ Since \(\log(x)\) is a concave function, we know that for any \(x\) in the interval \((k+1, k+2]\), \(\log(x)\) is greater than or equal to \(y\). Therefore, \(\log(k+2)>\frac{1}{k+1}+\log(k+1)\).

Combine Inequalities

Now, we have all the necessary inequalities to prove the inductive step. From Step 3, we have: $$ 1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{k}+\frac{1}{k+1}>\log(k+1)+\frac{1}{k+1} $$ And from Step 4, we have: $$ \log(k+2)>\frac{1}{k+1}+\log(k+1) $$ Combining these two inequalities, we get: $$ 1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{k}+\frac{1}{k+1}>\log(k+2) $$ Which is the inequality we wanted to prove for n=k+1. By the principle of mathematical induction, we have now proven that the inequality holds for all \(n \in \mathbb{N}\).

Key concepts

Proof Techniques

When facing a mathematical statement that needs verification for an infinite number of cases, one effective strategy is through mathematical induction. It's a proof technique based on a domino-effect principle.

Mathematical induction consists of two critical steps: the base case and the inductive step. Initially, the base case confirms the statement for the starting number, usually the smallest value, usually 1 in the context of natural numbers. This is akin to knocking down the first domino tile.

Then comes the inductive step, which comprises two sub-steps: assuming the statement is true for an arbitrary case 'k' (the induction hypothesis), and then showing that, under this assumption, the statement must also be true for the next case 'k+1'. If both these steps hold up, you've successfully proven the statement for all natural numbers, much like a continuous chain of falling dominoes.

It's crucial to note that this technique only proves the statement for natural numbers or cases that follow a similar sequential pattern. Mathematical induction is a favorite 'go-to' within discrete mathematics, particularly when dealing with sequences, series, and integer-based properties.

Natural Logarithm Properties

To fully understand the problem at hand, a grasp of the natural logarithm and its properties is pivotal. The natural logarithm, usually denoted as \( \log(x) \) or \( \ln(x) \) when the base is the special number e (approximately 2.718281828459), is a function that is crucial in various fields, including mathematics, physics, and engineering.

Some important properties are:

• The logarithm of a product is equal to the sum of the logarithms of its factors (i.e., \( \log(ab) = \log(a) + \log(b) \) ).
• The logarithm of a quotient is the difference of the logarithms (i.e., \( \log(\frac{a}{b}) = \log(a) - \log(b) \) ).
• The logarithm of a number raised to a power is the power times the logarithm of the number (i.e., \( \log(a^b) = b\log(a) \) ).

Moreover, the natural logarithm's graph is concave down, which means between any two points on the curve, the line segment connecting them lies below the curve, a property used in the aforementioned problem to establish a lower bound for \( \log(k+2) \) in order to facilitate the proof.

Sequences and Series

Our problem involves the sum of reciprocals, which is a harmonic series. In mathematics, a sequence is a list of numbers in a specific order, while a series is what you get if you sum the elements of a sequence. Different types of sequences and series have been studied, with their own sets of rules and behavior.

A well-known sequence is the arithmetic sequence, where each term after the first is obtained by adding a constant difference to the previous term. A geometric sequence is where each term is found by multiplying the previous one by a fixed non-zero number.

The harmonic series is particularly interesting because it grows very slowly - the sum of the reciprocals of the natural numbers diverges, meaning that as you add up more and more terms, the total sum grows without bound, but it does so at a pace that is slower than any other unbounded series with a simple rule of formation.

Understanding the behavior of different sequences and series is crucial in higher mathematics, as they are the bedrock of topics like calculus, number theory, and complex analysis. The study of series especially pertains to convergence and divergence, which inform us about the nature of infinite sums and their properties.