Problem 4 - Basic Analysis: Introduction to Real Analysis

Question

Use the geometric sum formula to show (for \(t \neq-1)\) $$ 1-t+t^{2}-\cdots+(-1)^{n} t^{n}=\frac{1}{1+t}-\frac{(-1)^{n+1} t^{n+1}}{1+t} $$ Using this fact show $$ \ln (1+x)=\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{n}}{n} $$ for all \(x \in(-1,1]\) (note that \(x=1\) is included). Finally, find the limit of the alternating harmonic series $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}=1-1 / 2+1 / 3-1 / 4+\cdots $$

Step-by-Step Solution

Answer

The alternating harmonic series converges to \(\ln(2)\).

Step 1: Apply Geometric Series Formula

The sum of a finite geometric series can be written as \(a + ar + ar^2 + \cdots + ar^n = \frac{a(r^{n+1} - 1)}{r-1} \). For the series \(1 - t + t^2 - \cdots + (-1)^n t^n\), we recognize this as a geometric series with first term \(a = 1\) and common ratio \(r = -t\). Hence, the sum is \(\frac{1 - (-t)^{n+1}}{1 + t}\).

Step 2: Simplify to Given Formula

From Step 1, the sum is \(\frac{1 - (-t)^{n+1}}{1 + t}\). Re-write \(1 - (-t)^{n+1}\) as \(1 + (-1)^{n+1} t^{n+1}\). The formula becomes \(\frac{1}{1+t} - \frac{(-1)^{n+1} t^{n+1}}{1+t}\), matching the given equation.

Step 3: Use Power Series Representation for Natural Logarithm

The series \(\ln(1+x) = \int_0^x \frac{1}{1+t} dt\). We expand \(\frac{1}{1+t}\) using the geometric series \(1 - t + t^2 - t^3 + \cdots\), valid for \(-1 < t < 1\). Thus, \(\int_0^x (1 - t + t^2 - t^3 + \cdots) dt = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}\).

Step 4: Conclude Limit of Alternating Harmonic Series

Plugging \(x=1\) into the logarithmic series from step 3, \(\ln(1+1) = \ln(2) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\). The conclusion is that the alternating harmonic series converges to \(\ln(2)\).

Key Concepts

Geometric Series

A geometric series is a series of terms where each term after the first is found by multiplying the previous term by a constant, known as the common ratio. For a geometric series with first term \(a\) and common ratio \(r\), the sum of the first \(n+1\) terms is given by the formula:

• \(a + ar + ar^2 + \cdots + ar^n = \frac{a(r^{n+1} - 1)}{r-1}\)

Understanding this formula is crucial, especially for cases where the common ratio satisfies \(-1 < r < 1\), ensuring convergence of the series to a finite sum. In the context of the problem, our specific series \(1 - t + t^2 - \cdots + (-1)^n t^n\) fits a geometric series pattern with a common ratio \(r = -t\). This allows us to find its sum effectively.
By substituting into the formula, the sum becomes \(\frac{1 - (-t)^{n+1}}{1+t}\), which can be further simplified into the given formula to explore deeper results, like expansions for other functions.

Natural Logarithm

The natural logarithm, \(\ln(x)\), is an important mathematical function often encountered in calculus. It is the inverse of the exponential function \(e^x\). For \(x > 0\), the natural logarithm can be defined in terms of an integral:

• \(\ln(x) = \int_1^x \frac{1}{t} \, dt\)

This means that \(\ln(1 + x) = \int_0^x \frac{1}{1+t} \, dt\) for \( -1 < x \leq 1\).
By using the power series representation of \(\frac{1}{1+t}\), expressed as \(1 - t + t^2 - t^3 + \cdots\), this integral can be expanded, resulting in the series \(x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\).
This expansion is critical because it provides a way to represent \(\ln(1+x)\) as an alternating power series, valid for the interval \(-1 < x \leq 1\). It also shows its practical use in computations involving logarithms over varying domains.

Alternating Series

An alternating series is a series where the signs of the terms alternate between positive and negative. Easily identified by terms like \(a_1 - a_2 + a_3 - a_4 + \cdots\), this kind of series might not always converge, but when it does, it conforms to the Alternating Series Test.

• A series \(\sum (-1)^{n+1} a_n\) converges if \(a_n\) is positive, decreasing, and approaches zero as \(n\) approaches infinity.

In the context of the exercise, the challenge involves understanding the convergence of the series \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\), the alternating harmonic series.
This series proves to converge to a finite value, specifically \(\ln(2)\). Understanding its convergence is powerful, offering insight into the values real-world alternating series can approach.

Power Series

A power series is an infinite sum of terms in the form \(a_n (x-c)^n\), representing functions over an interval of convergence. This mathematical tool is particularly useful in approximations and functional analysis.

• The power series for \(\ln(1+x)\) derived in the exercise is \(\sum_{n=1}^{\infty}\frac{(-1)^{n+1} x^n}{n} \), valid for \(-1 < x \leq 1\).

By representing \(\ln(1+x)\) using a power series, we gain a clear structure to approximate \(\ln\) over small values of \(x\).
This is essential when calculating logarithms in analysis or engineering where direct computation might be complex. Furthermore, this insight provides a strong foundation for exploring more advanced topics like Taylor series and convergence tests.