Problem 63 - A Course in Calculus and Real Analysis

Question

Let \(f:[1, \infty) \rightarrow \mathbb{R}\) be a nonnegative monotonically decreasing function. For \(n \in \mathbb{N}\), define \(c_{n}:=\sum_{k=1}^{n} f(k)-\int_{1}^{n} f(t) d t\). Show that \(\lim _{n \rightarrow \infty} c_{n}\) exists and $$ 0 \leq f(1)-\int_{1}^{2} f(t) d t \leq \lim _{n \rightarrow \infty} c_{n} \leq f(1) $$ Use this result to show that if $$ c_{n}:=1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n $$ then \(c_{n} \rightarrow \gamma\), where \(\gamma\) satisfies \(1-\ln 2<\gamma<1\).

Step-by-Step Solution

Answer

In summary, we showed that for a nonnegative monotonically decreasing function \(f\), the limit of the series \(c_n = \sum_{k=1}^{n}f(k) - \int_{1}^{n}f(t)dt\) exists. For the specific function, \(c_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \ln n\), we found that the limit converges to a constant \(\gamma\) such that \(1 - \ln 2 < \gamma < 1\).

Step 1: Observe the given series and integral

First, let's take a look at the series and the integral given in the exercise, the sum of the function \(f\): $$ c_{n} = \sum_{k=1}^{n} f(k) - \int_{1}^{n} f(t) dt $$ Notice that \(f\) is nonnegative and monotonically decreasing.

Step 2: Analyze the difference between the series and integral for successive terms

To better understand how to compare the series and integral, we can calculate the difference between each successive term: For \(k\geq 1\), we have $$ f(k) - \int_{k}^{k+1} f(t) dt. $$ Since \(f\) is monotonically decreasing and nonnegative, we can observe that $$ f(k+1) \leq \frac{1}{k+1} \int_{k}^{k+1} f(t) dt \leq f(k). $$

Step 3: Deduce the bounds for the difference between the series and the integral

Summing the inequalities over \(k\) from \(1\) to \(n\), we get $$ \sum_{k=1}^{n} f(k+1) \leq \sum_{k=1}^{n} \int_{k}^{k+1} f(t) dt \leq \sum_{k=1}^{n} f(k). $$ Since \(f(1)\) is nonnegative and finite, we can find the difference between the series and the integral within two bounds: $$ 0 \leq c_n \leq f(1) - \int_{1}^{2} f(t) dt. $$

Step 4: Prove that the limit of \(c_n\) exists

Since \(c_n\) is increasing (\(f(k)\) being nonnegative monotonically decreasing) and bounded above, the limit of \(c_n\) exists as \(n \rightarrow \infty\).

Step 5: Apply the result to the given function

Now we are asked to apply the result found in step 4 to the function: $$ c_{n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \ln n. $$ Here, \(f(k) = \frac{1}{k}\), which is nonnegative and monotonically decreasing on \([1, \infty)\). The integral of this function is $$ \int_{1}^{n} \frac{1}{t} dt = \ln n. $$ Using the result from step 4, we know that the limit of \(c_n\) exists and $$ f(1) - \int_{1}^{2} f(t)dt = 1 - \int_{1}^{2} \frac{1}{t} dt = 1 - \ln 2 \leq \lim_{n \rightarrow \infty} c_n \leq f(1) = 1. $$ Therefore, the sequence \(c_n\) converges to a constant \(\gamma\) such that \(1 - \ln 2 <\ \gamma < 1\).

Key Concepts

Monotonically Decreasing Function

When we talk about a **monotonically decreasing function**, we're referring to a function that consistently decreases, or at best, remains constant as it progresses. Essentially, if you pick any two points on the function, say \( x_1 \) and \( x_2 \) where \( x_1 < x_2 \), then it must be true that \( f(x_1) \geq f(x_2) \). This behavior ensures that the function doesn't increase as you move along the x-axis.
To illustrate with our given problem, consider \( f(k) = \frac{1}{k} \). For any \( k_1 < k_2 \), we have \( \frac{1}{k_1} > \frac{1}{k_2} \), making it a classic example of a monotonically decreasing function. This property is quite useful when working with sums and integrals.
Monotonicity is important because it helps establish bounds and convergence, particularly in calculus and analysis. It allows us to deduce certain behaviors for sequences and integrals, as it simplifies the estimation of function values over intervals.

Improper Integrals

An **improper integral** occurs when you're dealing with an integral that has an infinite limit or an integrand that approaches infinity at some point within the range of integration. These integrals often arise when calculating areas under curves that extend indefinitely.
Our problem deals with the integral from 1 to \( n \) of \( f(t) = \frac{1}{t} \), denoted by \( \int_{1}^{n} \frac{1}{t} dt \). As \( n \) approaches infinity, this integral tends toward \( \ln n \), reflecting the nature of the logarithmic growth. This growth is real and gradual, capable of dealing with limits that tend to infinity in a controlled manner.
Dealing with improper integrals requires careful handling because of their limits, often using techniques such as limit processes or variable substitution to evaluate them properly. The behavior of these integrals can vastly affect the convergence conclusions of related series or sums.

Series and Sequences

**Series and sequences** are fundamental concepts in calculus, often dealing with ordered lists of numbers. In this scenario, the series \( c_{n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \ln n \) is of particular interest. This is a partial sum known as the harmonic series, adjusted by the natural logarithm.
A sequence is simply a list of numbers arranged in a specific order. A series, on the other hand, involves summing elements of a sequence. In our problem, as \( n \) increases indefinitely, the sequence \( c_n \) converges to a specific value, \( \gamma \), known as the Euler-Mascheroni constant. This constant lies between 0 and 1, providing boundaries using known logs and sums.
Understanding series and sequences becomes vital as they dictate the behavior of convergence or divergence in various functions. By evaluating the limits of these sums and observing patterns, we can draw significant conclusions regarding their behavior over bounded or infinite intervals.