Question

Suppose \(a_{n}>0\) and \(\Sigma a_{n}\) converges. Put $$r_{n}=\sum_{m=n}^{\infty} a_{m} .$$ (a) Prove that $$\frac{a_{m}}{r_{m}}+\cdots+\frac{a_{n}}{r_{n}}>1-\frac{r_{n}}{r_{m}}$$ if \(m

Short answer

Question: Prove that if the series \(\sum_{n=1}^\infty a_{n}\) converges, and each term \(a_{n} = r_{n} - r_{n+1}\) for \(r_{n} > 0, n = 1, 2, \dots\), then (a) the series \(\sum_{n=1}^\infty \frac{a_{n}}{r_{n}}\) diverges and (b) the series \(\sum_{n=1}^\infty \frac{a_{n}}{\sqrt{r}_n}\) converges. Answer: The result can be proven by first showing that the inequality \(\frac{a_{n}}{r_{n}} > 1 - \frac{r_n}{r_m}\) for \(m < n\) holds true for all \(n\). This will help deduce that the series \(\sum_{n=1}^\infty \frac{a_{n}}{r_{n}}\) diverges. Next, we can prove the inequality \(\frac{a_{n}}{\sqrt{r_n}} < 2(\sqrt{r_n} - \sqrt{r_{n+1}})\) for all \(n\). This inequality can then be used to show that the series \(\sum_{n=1}^\infty \frac{a_{n}}{\sqrt{r}_n}\) converges using the comparison test.

Step-by-step solution

Prove the inequality for \(m

Since \(\Sigma a_{n}\) converges, we know that \(a_{n}\to 0\) as \(n \to \infty\). Therefore, we have \(r_{n+1}n-m+1-\frac{r_{n+1}}{r_{m}}>1-\frac{r_{n+1}}{r_{m}}>1-\frac{r_{n}}{r_{m}}$$ The inequality holds for any \(m<n\).

Deduce that \(\sum\frac{a_{n}}{r_{n}}\) diverges#

Consider the subsequence of partial sums \(S_{k}=\frac{a_{1}}{r_{1}}+\cdots+\frac{a_{2^{k}}}{r_{2^{k}}}\). We have $$S_{k}-S_{k-1}=\frac{a_{2^{k-1}+1}}{r_{2^{k-1}+1}}+\cdots+\frac{a_{2^{k}}}{r_{2^{k}}}>1-\frac{r_{2^{k}}}{r_{2^{k-1}+1}}\geq 1-\frac{r_{2^{k}}}{r_{2^{k-1}}}$$. Therefore, \(S_{k}>1-\frac{1}{2^{k-1}}\) when summed from \(k=1\) to infinity, so \(\lim_{k\to\infty} S_{k}\) is infinity. Hence, the series \(\sum\frac{a_{n}}{r_{n}}\) diverges. ## Part (b) ## We will now prove the given inequality, i.e., \(\frac{a_{n}}{\sqrt{r_{n}}} < 2(\sqrt_r_n - \sqrt{r_{n+1}})\), and then deduce that the series \(\sum \frac{a_{n}}{\sqrt{r}_{n}}\) converges.

Prove the inequality for all \(n\)#

As mentioned before, we already know that \(a_{n}=r_{n}-r_{n+1}\), so \(\frac{a_{n}}{\sqrt{r_{n}}}<\frac{r_{n}-r_{n+1}}{\sqrt{r_{n}}}\). Now, consider the function \(f(x)=\sqrt{x}\), with \(x\geq 0\). The second derivative \(f''(x)=\frac{-1}{4x^{3/2}}\), which is always negative for \(x>0\). Therefore, \(\sqrt{x}\) is a concave function for all \(x>0\). Hence, by Jensen's inequality, we get: $$\frac{\sqrt{r_n}-\sqrt{r_{n+1}}}{r_n-r_{n+1}}>\frac{1}{2}(f'(r_n)+f'(r_{n+1}))=\frac{1}{2(\sqrt{r_n}+\sqrt{r_{n+1}})}$$ Multiplying this inequality with \(2(r_n-r_{n+1})\) gives $$2(\sqrt{r_n}-\sqrt{r_{n+1}})>\frac{r_n-r_{n+1}}{\sqrt{r_n}+\sqrt{r_{n+1}}}\geq\frac{a_n}{\sqrt{r_n}}$$ which is the required inequality.

Deduce that \(\sum\frac{a_{n}}{\sqrt{r}_{n}}\) converges#

Since we have shown that \(\frac{a_{n}}{\sqrt{r}_{n}}<2(\sqrt{r_{n}}-\sqrt{r_{n+1}})\), we can use the comparison test for convergence. Consider the series \(\sum 2(\sqrt{r_{n}}-\sqrt{r_{n+1}})=2(\sqrt{r_{1}}-\sqrt{r_{2}})+2(\sqrt{r_{2}}-\sqrt{r_{3}})+\cdots\). The terms cancel out, and we get \(\sum 2(\sqrt{r_{n}}-\sqrt{r_{n+1}})=2(\sqrt{r_{1}}-\sqrt{r_{\infty}})=2\sqrt{r_{1}}\), which is a convergent series since \(r_1\) is finite. Thus, by the comparison test, the series \(\sum\frac{a_{n}}{\sqrt{r}_{n}}\) also converges.

Key concepts

Convergence and Divergence

Understanding the concepts of convergence and divergence is fundamental in mathematical analysis, particularly when dealing with sequences and series. A sequence of numbers is simply a list of numbers in a specific order, while a series is the sum of a sequence's terms. But not all series neatly sum up to a finite value; they can either converge or diverge.

A series converges if the sum of its terms approaches a specific value as more and more terms are added. In the context of the given exercise, the series \(\Sigma a_{n}\) is said to converge because as we add terms, the series reaches closer to a particular limit. For a series to converge, the terms \(a_n\) must generally get smaller; in fact, they should become infinitesimal as \(n\) grows, approaching zero.

On the other hand, a series diverges if its sum does not approach a specific value as terms are added. This can occur if the terms do not decrease rapidly enough or even start increasing at some point. In the exercise's part (a), we see a brilliant example where the comparison between terms \(\frac{a_{n}}{r_{n}}\) leads to a proof that the series \(\sum \frac{a_{n}}{r_{n}}\) diverges because the partial sums grow without bound.

Sequence and Series

In mathematical analysis, sequences and series provide the groundwork for understanding many concepts, including functions, integrals, and more. A sequence is an ordered list of objects or numbers that may follow a particular pattern or rule. The notation \(a_n\) is commonly used to denote the \(n\)th term of a sequence, where \(n\) is typically a positive integer.

Series, built from sequences, are the summation of the terms of a sequence. When we talk about the sum of a series, like \(\Sigma a_{n}\) or \(r_n\) in the exercise, we are discussing the result of adding all the terms of a sequence together, potentially up to infinity. Understanding the behavior of these additions is critical in determining their convergence or divergence. An important aspect of series used in many proofs, including the exercise, is that they can be manipulated algebraically much like finite sums.

Inequalities in Analysis

Inequalities play a crucial role in mathematical analysis as they help establish the boundaries within which functions or sequences operate. They are essential for proving various theorems about limits, convergence, and divergence. The initial part of the exercise uses an inequality to show that a certain series diverges.

Inequalities are not just about less than or greater than relationships; they can compare rates of growth, magnitudes of decay, and much more. The exercise's step 1 establishes a fundamental inequality which paves the way for proving that \(\sum \frac{a_{n}}{r_{n}}\) diverges. Similarly, part (b) of the problem shows another inequality involving square roots that leads to the proof that \(\sum \frac{a_{n}}{\sqrt{r}_{n}}\) converges. These inequalities help in summarizing the behavior of sequences and series without the need for specific computations for each term, which might be impossible for infinitely long series.