Question

In each of the following problems, use the estimate \(\left|R_{N}\right| \leq b_{N+1}\) to find a value of \(N\) that guarantees that the sum of the first \(N\) terms of the alternating series \(\sum_{n=1}^{\infty}(-1)^{n+1} b_{n}\) differs from the infinite sum by at most the given error. Calculate the partial sum \(S_{N}\) for this \(N\).[T] \(b_{n}=\ln \left(1+\frac{1}{n}\right)\), error \(<10^{-3}\)

Short answer

For an error less than \(10^{-3}\), use \(N=1000\). Calculate \(S_{1000}\) by evaluating the sum of the first 1000 terms.

Step-by-step solution

Understand the Remainder Estimate

The remainder estimate for an alternating series is given by \(|R_N| \leq b_{N+1}\). This tells us that the error from truncating the series after \(N\) terms is at most the absolute value of the next term.

Identify the Terms in the Series

We are given \(b_n = \ln \left(1+\frac{1}{n}\right)\). This implies the series in question is \(\sum_{n=1}^{\infty}(-1)^{n+1} \ln \left(1+\frac{1}{n}\right)\).

Set Up the Inequality for the Remainder Estimate

To ensure the error is less than \(10^{-3}\), we need \( \ln \left(1+\frac{1}{N+1}\right) < 10^{-3} \).

Analyze the Natural Logarithm Term

We know that \(\ln(1+x) \approx x\) when \(x\) is small. Hence, \(\ln \left(1+\frac{1}{N+1}\right) \approx \frac{1}{N+1}\).

Solve for N

Solve the inequality \(\frac{1}{N+1} < 10^{-3}\) to find the smallest integer \(N\).\[ \frac{1}{N+1} < 10^{-3} \Rightarrow N+1 > 1000 \Rightarrow N > 999 \]Thus, \(N = 1000\).

Calculate the Partial Sum \(S_N\) for \(N = 1000\)

The partial sum \(S_{1000}\) is given by:\[ S_{1000} = \sum_{n=1}^{1000}(-1)^{n+1} \ln \left(1+\frac{1}{n}\right) \]Calculate this sum by adding terms sequentially from \(n=1\) to \(n=1000\), alternating the signs.

Key concepts

Remainder Estimate

The remainder estimate is a core concept used to determine how close a partial sum of an infinite series is to the actual total of that series. For alternating series, a special kind of series where each term alternates in sign, the remainder is estimated using the formula:

• \(|R_N| \leq b_{N+1}\)

This inequality tells us that the difference between the infinite series and its partial sum after \(N\) terms is at most the absolute value of the next term in the sequence.
This method is particularly useful for approximating the sum of an infinite series to a certain level of precision, helping us decide how many terms \(N\) we need to sum before stopping. In practical terms, this helps provide a manageable solution for series that would otherwise require us to compute an infinite amount of terms.
For the problem at hand, we used this idea to ensure that the series' sum is accurate to within an error of less than \(10^{-3}\). This is achieved by selecting \(N\) such that

• \(\left|R_{N}\right| \leq 10^{-3}\)

allowing us to find that \(N = 1000\) guarantees the desired level of precision.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is a logarithm whose base is the mathematical constant \(e\), approximately equal to 2.71828. It's a fundamental function in calculus and mathematical analysis, frequently used to model exponential growth and decay processes.
For small values, the natural logarithm can be approximated using the formula:

• \( \ln(1 + x) \approx x \)

This is particularly helpful in our exercise, as it allows us to derive a simple approximation for \( \ln\left(1 + \frac{1}{N+1}\right) \) when \(N\) becomes large. By recognizing \( \ln\left(1 + \frac{1}{n}\right) \) in the series, this approximation simplifies the calculation of the term \(b_{n}\) to \(\frac{1}{n}\) when we are trying to establish the remainder estimate.
This simplification means that the error term \( \ln\left(1 + \frac{1}{N+1}\right) < 10^{-3} \) can instead be treated as \( \frac{1}{N+1} < 10^{-3} \), making the task of calculating \(N\) much more feasible.

Partial Sum

A partial sum refers to the sum of the first \(N\) terms in an infinite series. Unlike an infinite sum, a partial sum is a finite value and thus much more manageable in calculations. In the context of alternating series, the terms alternate in sign, which often results in a partial sum that converges towards the series’ actual infinite sum.
Calculating the partial sum \(S_N\) is straightforward:

• \( S_{N} = \sum_{n=1}^{N} (-1)^{n+1} b_{n} \)

For our specific problem, you would compute the sum by adding and subtracting the terms in the series up to \(n = 1000\), ensuring to alternate the signs. This alternating behavior is because of the factor \((-1)^{n+1}\), which results in subtracting every even term and adding every odd term.
The goal of finding \(S_{1000}\) is to approximate the infinite series sum with a limited error, as determined by our remainder estimate. This computation provides a practical way to understand the progression and behavior of the series without the need for summing infinitely many terms.