Problem 3 - The Calculus with Analytic Geometry

Question

Find the first four elements of the sequence of partial sums \(\left\\{s_{n}\right\\}\) and find a formula for \(s_{n}\) in terms of \(n ;\) also, determine if the infinite series is convergent or divergent, and if it is convergent, find its sum.\(\sum_{n=1}^{+\infty} \ln \frac{n}{n+1}\)

Step-by-Step Solution

Answer

The first four elements are \(-\ln 2, -\ln 3, -\ln 4, -\ln 5\) and the infinite series is divergent.

Step 1 - Understand the given sequence

Identify the sequence given in the problem. The given sequence is \(\sum_{n=1}^{+\infty} \ln \frac{n}{n+1}\).

Step 2 - Rewrite the given term

Rewrite the term \(\ln \frac{n}{n+1}\) using properties of logarithms. We can use the property \(\ln \frac{a}{b} = \ln a - \ln b\). This gives us: \[\ln \frac{n}{n+1} = \ln n - \ln (n+1)\].

Step 3 - Write the partial sum

Express the partial sum \(s_n\) of the series by summing up to the first \(n\) terms. This gives us: \[s_n = \sum_{k=1}^{n} \left( \ln k - \ln (k+1) \right)\].

Step 4 - Simplify the partial sum

Observe the pattern in the sum to simplify it. Notice that most terms cancel out, leaving: \[s_n = \ln 1 - \ln (n+1)\]. Since \(\ln 1 = 0\), it simplifies further to: \[s_n = -\ln (n+1)\].

Step 5 - First four elements of the sequence

Compute the first four elements of the sequence \left\{s_{n}\right\}\. For \(n = 1\): \(s_1 = -\ln 2\). For \(n = 2\): \(s_2 = -\ln 3\). For \(n = 3\): \(s_3 = -\ln 4\). For \(n = 4\): \(s_4 = -\ln 5\). Thus, the first four elements are: \(-\ln 2, -\ln 3, -\ln 4, -\ln 5\).

Step 6 - Determine convergence or divergence

Examine the terms \(s_n = -\ln (n+1)\) as \(n \to +\infty\). As \(n\) grows larger, \(-\ln (n+1)\) approaches \(-\infty\). Therefore, the infinite series is divergent because \(s_n\) does not converge to a finite value.

Key Concepts

Sequence of Partial Sums

To understand the given sequence, we start by identifying how to form the sequence of partial sums. The original infinite series we have is \(\sum_{n=1}^{+\infty} \ln \frac{n}{n+1}\). Using properties of logarithms, this can be rewritten as a telescoping series:\(\sum_{n=1}^{+\infty} \left( \ln n - \ln (n+1) \right)\). To compute the partial sums, take the sum up to the first \(n\) terms:\(\sum_{k=1}^{n} \left( \ln k - \ln (k+1) \right)\). By simplifying, we observe that many terms cancel out, leaving us with:\(\ln 1 - \ln (n+1)= -\ln (n+1) \). This is the formula for \(s_n\) in terms of \(n\). The first four elements of this sequence are:\(-\ln 2, -\ln 3, -\ln 4, -\ln 5\). This pattern shows how the sequence evolves and gives us insights for further analysis.

Convergence and Divergence

A critical aspect of analyzing infinite series is determining whether they converge or diverge. For our series, we look at the behavior of the partial sums \(s_n = -\ln (n+1)\) as \(n \to +\infty\). If \(s_n\) approaches a finite limit as \(n\) increases indefinitely, the series converges. However, as \(n\) grows larger, \(-\ln (n+1)\) continues to decrease without bound, approaching \(-\infty\). Since the sequence of partial sums does not approach a finite number, we conclude that the infinite series is divergent. Understanding this helps in determining the series' behavior over long terms.

Logarithmic Properties

Logarithmic properties often simplify complex series, as seen with the series \(\sum_{n=1}^{+\infty} \ln \frac{n}{n+1}\). By applying the property \(\ln \frac{a}{b} = \ln a - \ln b\), we transformed the term \(\ln \frac{n}{n+1}\) into \(\ln n - \ln (n+1)\). This transformation is crucial as it reveals the telescoping nature of the series, where many intermediate terms cancel out. Such cancellations simplify computation, leading to more manageable expressions for partial sums. These logarithmic transformations are powerful tools in series analysis, saving considerable effort in deriving results.

Series Simplification

Simplifying series helps in determining their properties more easily. In our example, by rewriting \(\ln \frac{n}{n+1}\) as \(\ln n - \ln (n+1)\) and summing the terms, most intermediate terms cancel out, which is characteristic of a telescoping series. This cancellation simplifies the partial sum to \(s_n = -\ln (n+1)\). Simplifying the series not only makes calculation straightforward but also provides clear insights into the series' behavior. For example, in determining convergence or divergence, the simplified form immediately suggests the tendency of the sequence's terms and thus its series.