Question

Suppose \(a_{n}>0, s_{n}=a_{1}+\cdots+a_{n}\), and \(\Sigma a_{n}\) diverges. (a) Prove that \(\sum \frac{a_{n}}{1+a_{n}}\) diverges. (b) Prove that $$\frac{a_{N+1}}{s_{N+1}}+\cdots+\frac{a_{N+k}}{s_{N+k}} \geq 1-\frac{s_{N}}{s_{N+k}}$$ and deduce that \(\sum \frac{a_{n}}{s_{n}}\) diverges. (c) Prove that $$ \frac{a_{n}}{s_{n}^{2}} \leq \frac{1}{s_{n-1}}-\frac{1}{s_{n}}$$ and deduce that \(\sum \frac{a_{n}}{s_{n}^{2}}\) converges. (d) What can be said about $$ \sum \frac{a_{n}}{1+n a_{n}} \text { and } \sum \frac{a_{n}}{1+n^{2} a_{n}} ? $$

Short answer

Based on the given information and step by step solution, allow for the following short answer: The series \(\sum\frac{a_n}{1+a_n}\) diverges, as proven in part (a). The series \(\sum\frac{a_n}{s_n}\) diverges as well, due to the proven inequality in part (b). In part (c), it is shown that the series \(\sum\frac{a_n}{s_n^2}\) converges because the inequality \(\frac{a_n}{s_n^2} \leq \frac{1}{s_{n-1}}-\frac{1}{s_n}\) is true and the right-hand side series converges. Lastly, the convergence of the series \(\sum\frac{a_n}{1+n^2a_n}\) cannot be determined with the provided information.

Step-by-step solution

Compare series with harmonic series

We know that the harmonic series \(\sum \frac{1}{n}\) diverges. Observe that for each \(n\), \(\frac{a_n}{1+a_n}\ge\frac{a_n}{1+n a_n}\), because \({a_n}>0\) and increasing the denominator decreases the fraction value. We will show that the series \(\sum \frac{a_n}{1+n a_n}\) diverges and deduce that the series \(\sum\frac{a_n}{1+a_n}\) also diverges.

Use comparison test

Let \(b_n = \frac{1}{n}\), so the series \(B = \sum b_n\) is the harmonic series. By the comparison test, if we can show that \(\frac{a_n}{1+na_n} \geq b_n\) for all \(n\), we can conclude that \(\Sigma \frac{a_n}{1+n a_n}\) diverges. Notice that \(\frac{a_n}{1+na_n} \geq b_n\) is equivalent to \(a_n \ge n b_n = \frac{n}{n}\), which is true because \(a_n>0\) for all \(n\). So by the comparison test, the series \(\Sigma \frac{a_n}{1+n a_n}\) diverges, and it follows that the series \(\Sigma \frac{a_n}{1+a_n}\) also diverges. (b) Prove the inequality \(\frac{a_{N+1}}{s_{N+1}}+\cdots+\frac{a_{N+k}}{s_{N+k}} \geq 1-\frac{s_{N}}{s_{N+k}}\) and deduce that the series \(\sum\frac{a_n}{s_n}\) diverges.

Deduce inequality

Notice that for \(1\le j\le k\), \(s_j > 0\) (since \(a_i>0\)) and \(s_{N+j} > s_N\). Then we can write the inequality as: $$ \frac{a_{N+1}}{s_{N+1}}+\cdots+\frac{a_{N+k}}{s_{N+k}} \ge \frac{a_{N+1}}{s_{N+k}}+\cdots+\frac{a_{N+k}}{s_{N+k}} = 1-\frac{s_N}{s_{N+k}} $$ The inequality is now proven.

Divergence of series

To show that \(\sum\frac{a_n}{s_n}\) diverges, consider taking the partial sum up to index \(n = N+k\): $$ \sum_{n=1}^{N+k}\frac{a_n}{s_n} = \sum_{n=1}^N\frac{a_n}{s_n} + \sum_{n=N+1}^{N+k}\frac{a_n}{s_n} \ge \sum_{n=1}^N\frac{a_n}{s_n} + 1-\frac{s_N}{s_{N+k}} $$ Here, to make the partial sum arbitrarily large, we can choose \(N\) large enough so that the series \(\sum \frac{a_n}{s_n}\) diverges. (c) Prove the inequality \(\frac{a_n}{s_n^2} \leq \frac{1}{s_{n-1}}-\frac{1}{s_n}\) and deduce that the series \(\sum\frac{a_n}{s_n^2}\) converges.

Show inequality

To show the inequality, \(\frac{a_n}{s_n^2} \leq \frac{1}{s_{n-1}}-\frac{1}{s_n}\), we can rewrite the right-hand side as \(\frac{s_n-s_{n-1}}{s_{n-1}s_n}\). Then the inequality becomes \(a_n s_{n-1} \leq s_n - s_{n-1}\). Since \(s_n = s_{n-1} + a_n\), this inequality holds.

Convergence of series

Now, consider the series \(\sum\frac{1}{s_{n-1}}-\frac{1}{s_n}\). This is a telescoping series, as the terms cancel out to leave just \(\frac{1}{s_0}-\frac{1}{s_n}\), which converges, because \(\lim_{n \to \infty}\frac{1}{s_n}=0\). Since the inequality \(\frac{a_n}{s_n^2} \le \frac{1}{s_{n-1}}-\frac{1}{s_n}\), it follows that the series \(\sum\frac{a_n}{s_n^2}\) also converges, by the comparison test. (d) Determine the convergence of the series \(\sum\frac{a_n}{1+na_n}\) and \(\sum\frac{a_n}{1+n^2a_n}\).

Recall part (a)

We found in part (a) that the series \(\sum\frac{a_n}{1+na_n}\) diverges.

Convergence of second series

To determine the convergence of the series \(\sum\frac{a_n}{1+n^2a_n}\), we can use a similar technique as in part (a) by comparing it with the harmonic series. Note that for each \(n\), \(\frac{a_n}{1+n^2a_n}\le\frac{a_n}{1+n a_n}\). Since the series \(\sum\frac{a_n}{1+na_n}\) diverges, we cannot conclude the convergence of the series \(\sum\frac{a_n}{1+n^2a_n}\) based on this comparison. Further analysis is required to determine its convergence or divergence. In conclusion, the series \(\sum\frac{a_n}{1+na_n}\) diverges, and the convergence of the series \(\sum\frac{a_n}{1+n^2a_n}\) cannot be determined with the provided information.

Key concepts

Harmonic Series

The harmonic series is a well-known and fundamental concept in mathematical analysis. It is defined as the infinite sum \( \sum \frac{1}{n} \) where \( n \) is a positive integer. This series, despite being a simple sum of reciprocals of natural numbers, is divergent. This means that its sum is infinite and never approaches a fixed finite value even if you add infinitely many terms.

This property of divergence is crucial because it demonstrates that not all infinite series have finite sums. The harmonic series serves as a benchmark in the analysis of many other series. When comparing another series to the harmonic series, you can often determine whether the other series converges (sum approaches a finite limit) or diverges.

In the context of this exercise, the harmonic series acts as a point of comparison to evaluate the behavior of the series defined by \( \sum \frac{a_n}{1+a_n} \) and \( \sum \frac{a_n}{1+na_n} \). Since the harmonic series diverges, if these series can be shown to mimic its behavior, we can infer that they also diverge.

Comparison Test

The comparison test is a valuable tool in determining the convergence or divergence of infinite series. By comparing a given series with a second series whose behavior (convergence or divergence) is already known, you can make inferences about the first series.

The idea is straightforward. If you have a series \( \sum a_n \) and you've identified a well-understood benchmark series \( \sum b_n \) which is easier to analyze, you compare the terms of the two series. If \( a_n \geq b_n \) for all \( n \) and the series \( \sum b_n \) diverges, then \( \sum a_n \) also diverges. Conversely, if \( a_n \leq b_n \) for all \( n \) and the series \( \sum b_n \) converges, then \( \sum a_n \) also converges.

In this exercise, the series \( \sum \frac{a_n}{1+na_n} \) is compared to the harmonic series \( \sum \frac{1}{n} \). Since the harmonic series diverges, and the terms of the given series are sufficiently large by the inequality \( \frac{a_n}{1+na_n} \geq \frac{1}{n} \), the comparison test confirms that \( \sum \frac{a_n}{1+na_n} \) also diverges. This result helps deduce the behavior of other related series in the exercise.

Convergence and Divergence

Convergence and divergence are key concepts when working with series. A series converges if the sequence of its partial sums approaches a specific finite value as the number of terms goes to infinity. If this sequence doesn't approach any particular value, the series diverges.

Understanding the behavior of series in terms of convergence or divergence is crucial, especially when analyzing the limits, sums, or other properties of functions in calculus and real analysis.

For instance, in this problem, we analyze a variety of series: \( \sum \frac{a_n}{1+a_n} \) and \( \sum \frac{a_n}{1+na_n} \) are shown to diverge by comparing them to the divergent harmonic series. However, using a different approach, the series \( \sum \frac{a_n}{s_n^2} \) is shown to converge. This contrast highlights the importance of method selection and the properties of series terms for determining whether a series converges or diverges. Other factors like rearrangement or specific tests (like the telescoping series) can also influence conclusions about series convergence or divergence.