Question

If \(\left\\{s_{k}\right\\}\) is a complex sequence, define its arithmetic means \(\sigma_{n}\) by $$\sigma_{n}=\frac{s_{0}+s_{1}+\cdots+s_{n}}{n+1} \quad(n=0,1,2, \ldots) .$$ (a) If \(\lim s_{n}=s\), prove that \(\lim \sigma_{n}=s\). (b) Construct a sequence \(\left\\{s_{n}\right\\}\) which does not converge, although \(\lim \sigma_{n}=0 .\) (c) Can it happen that \(s_{n}>0\) for all \(n\) and that lim sup \(s_{n}=\infty\), although lim \(\sigma_{n}=0\) ? ( \(d\) ) Put \(a_{n}=s_{n}-s_{n-1}\), for \(n \geq 1\). Show that $$s_{n}-\sigma_{n}=\frac{1}{n+1} \sum_{k=1}^{n} k a_{k} .$$ Assume that \(\lim \left(n a_{n}\right)=0\) and that \(\left\\{\sigma_{n}\right\\}\) converges. Prove that \(\left\\{s_{n}\right\\}\) converges. [This gives a converse of \((a)\), but under the additional assumption that \(\left.n a_{n} \rightarrow 0 .\right]\) (e) Derive the last conclusion from a weaker hypothesis: Assume \(M<\infty\), \(\left|n a_{n}\right| \leq M\) for all \(n\), and \(\lim \sigma_{n}=\sigma .\) Prove that \(\lim s_{n}=\sigma\), by completing the following outline: If \(m0\) and associate with each \(n\) the integer \(m\) that satisfies $$m \leq \frac{n-\varepsilon}{1+\varepsilon}

Short answer

(a) Prove that if \(\lim s_{n}=s\), then \(\lim \sigma_{n}=s\). The proof is as follows: We know that the definition of \(\sigma_n\) is $$\sigma_{n}=\frac{s_{0}+s_{1}+\cdots+s_{n}}{n+1}$$ Given that \(\lim\limits_{n\rightarrow\infty} s_n = s\), we have: $$\lim_{n\rightarrow\infty}\sigma_n=\lim_{n\rightarrow\infty}\left( \frac{s}{n+1}\right)$$ The limit of the expression above is equal to \(s\). So, we can conclude that if \(\lim_{n\rightarrow\infty} s_n = s\), then \(\lim_{n\rightarrow\infty} \sigma_n = s\). (b) Construct a sequence \(\left\\{s_{n}\right\\}\) that does not converge but has \(\lim \sigma_{n}=0\). The example sequence \(\left\\{s_n\right\\}\) is as follows: $$s_n=(-1)^n$$ This sequence does not converge, since the terms oscillate between 1 and -1. However, the arithmetic means \(\sigma_n\) tend to 0 as \(n \rightarrow \infty\). (c) Check if a sequence \(s_{n}>0\) can exist such that lim sup \(s_{n}=\infty\) and lim \(\sigma_{n}=0\) simultaneously. The answer is no; it's not possible to have a sequence with the given conditions. (d) Investigate the relationship between \(a_n=s_{n}-s_{n-1}\) and prove some properties related to the convergence. One of the relations is \(s_n-\sigma_n=\frac{1}{n+1}\sum_{k=1}^{n} ka_k\). Given the conditions such as \(\lim (n a_n) = 0\) and \(\left\\{\sigma_n\right\\}\) converges, we can conclude that \(\left\\{s_n\right\\}\) converges as well. (e) Prove that if certain conditions are satisfied, then \(\lim s_n = \sigma\). Under the weaker hypothesis, our desired claim holds: if \(\lim \sigma_n=\sigma\) and the provided conditions are satisfied, then \(\lim s_n=\sigma\).

Step-by-step solution

(a) Proving that \(\lim \sigma_{n}=s\) given \(\lim s_{n}=s\)

Given that \(\lim\limits_{n\rightarrow\infty} s_n = s\), we need to show that \(\lim\limits_{n\rightarrow\infty} \sigma_n = s\). We know that the definition of \(\sigma_n\) is $$\sigma_{n}=\frac{s_{0}+s_{1}+\cdots+s_{n}}{n+1}$$ Let's find the limit of \(\sigma_n\) as n approaches infinity. $$\lim_{n\rightarrow\infty}\sigma_n=\lim_{n\rightarrow\infty}\left( \frac{s_{0}+s_{1}+\cdots+s_{n}}{n+1}\right)$$ Using the limit properties, we can find the limit: $$\lim_{n\rightarrow\infty}\sigma_n=\lim_{n\rightarrow\infty}\left( \frac{s}{n+1}\right)$$ Since \(\lim_{n\rightarrow\infty} s_n = s\), the whole expression becomes: $$\lim_{n\rightarrow\infty} \sigma_n = s$$ So, we can conclude that if \(\lim_{n\rightarrow\infty} s_n = s\), then \(\lim_{n\rightarrow\infty} \sigma_n = s\).

(b) Constructing a sequence \(\left\\{s_n\right\\}\) with \(\lim \sigma_{n}=0\) but non-convergent

We can define a sequence \(\left\\{s_n\right\\}\) as follows: $$s_n=(-1)^n$$ So, the arithmetic means \(\sigma_n\) are: $$\sigma_n=\frac{s_{0}+s_{1}+\cdots+s_{n}}{n+1}=\frac{1+(-1)+1+(-1)+ \cdots + (-1)^n}{n+1}$$ Let's calculate the limit of this sequence for when \(n \rightarrow \infty\). We have: $$\lim_{n\rightarrow\infty} \sigma_n = 0$$ However, the sequence, \(\left\\{s_n\right\\} = \{1,-1,1,-1, \cdots\}\), does not converge, since the terms oscillate between 1 and -1. Therefore, the constructed sequence satisfies the given criteria.

(c) Investigating the existence of a sequence satisfying the given conditions

In order to prove the existence of a sequence \(s_n>0\) with lim sup \(s_n = \infty\) and lim \(\sigma_n = 0\), we can construct the following sequence: \(s_n=\frac{1}{n}\) for \(n=1,2,\dots,\) and steps we can have the limit of \(\sigma_n\). By calculating the arithmetic means: $$\sigma_n=\frac{\frac{1}{1}+\frac{1}{2}+ \cdots + \frac{1}{n}}{n+1}$$ This is a harmonic sequence which is a well-known divergent sequence. However, the limit of the arithmetic means \(\lim\limits_{n\rightarrow\infty} \sigma_n = 0\). Now we need to check for lim sup \(s_n\). We have: $$\limsup_{n\rightarrow\infty} s_n = \limsup_{n\rightarrow\infty} \frac{1}{n} = 0$$ Since the lim sup \(s_n = 0\), it contradicts the given condition lim sup \(s_n = \infty\). Thus, it's not possible to have a sequence with the given conditions.

(d) Relating \(a_n\) and \(\sigma_n\) and finding the convergence of \(\left\\{s_n\right\\}\)

Given \(a_n=s_n-s_{n-1}\) for \(n \geq 1\), we need to show that: $$s_n-\sigma_n=\frac{1}{n+1}\sum_{k=1}^{n} ka_k$$ Let's simplify the left side of the equation: $$s_n-(\sigma_n)=s_n-\frac{s_{0}+s_{1}+\cdots+s_{n}}{n+1}$$ Therefore, \(s_n(n+1) - (s_0 + s_1 + \dots + s_n)=(n+1)s_n-(s_0+s_1+ \cdots +s_n)\) Let's rewrite the sum \(s_0 + \dots +s_n\), (substituting \(s_k = a_k + s_{k-1}\), \(k \ge 1\)): $$s_n-\sigma_n= (1\cdot a_1) + (2\cdot a_2) + (3\cdot a_3) + \cdots + (n\cdot a_n)-\frac{1}{n+1}$$ Thus, we've proved the stated relationship. Next, assume: 1. \(\lim (n a_n) = 0\) 2. \(\left\\{\sigma_n\right\\}\) converges We need to show that \(\left\\{s_n\right\\}\) converges as well. Let \(s = \lim \sigma_n\). Then, for a given \(\epsilon > 0\), there exists \(N \in \mathbb{N}\) such that \(|\sigma_n - s| < \frac{\epsilon}{2}\) for all \(n \ge N\). Also, we have: $$\left |s_n-\sigma_n\right|=\left|\frac{1}{n+1}\sum_{k=1}^{n} ka_k\right| \leq \frac{1}{n+1}\sum_{k=1}^{n} |k a_k|$$ Given \(\lim(n a_n) = 0\), for a given \(\epsilon > 0\), there exists \(N_1 \in \mathbb{N}\) such that \(|n a_n| < \frac{\epsilon}{2}\) for all \(n \ge N_1\). So, for \(n > \text{max}\{N, N_1\}\), we have: $$\left |s_n-\sigma_n\right| \leq \frac{1}{n+1}\sum_{k=1}^{n} |k a_k| \leq \frac{\epsilon}{2}$$ Hence, \(|s_n - s| = |(s_n - \sigma_n) - (\sigma_n - s)| \leq |s_n - \sigma_n| + |\sigma_n - s| < \epsilon\) for all \(n \ge \text{max}\{N, N_1\}\). We can conclude that \(\lim s_n = s\), so \(\left\\{s_n\right\\}\) converges.

(e) Proving convergence with the given weaker hypothesis

Given: 1. \(M<\infty\) 2. \(|\left.n a_{n}\right| \leq M\) for all \(n\) 3. \(\lim \sigma_{n}=\sigma\) We are given the following identity: $$s_{n}-\sigma_{n}=\frac{m+1}{n-m}\left(\sigma_{n}-\sigma_{m}\right)+\frac{1}{n-m}\sum_{t=m+1}^{n}\left(s_{n}-s_{i}\right)$$ Start by using the provided information and inequalities: $$\left|s_{n}-s_{i}\right| \leq \frac{(n-i) M}{i+1} \leq \frac{(n-m-1) M}{m+2}$$ Fix an \(\epsilon > 0\), and associate with each \(n\) the integer \(m\) that satisfies: $$m \leq \frac{n-\epsilon}{1+\epsilon}<m+1$$ By our hypothesis, \((m+1) /(n-m) \leq 1 / \epsilon\) and \(\left|s_{n}-s_{i}\right|<M \epsilon\). Using these inequality, we can deduce: $$\lim _{n \rightarrow \infty} \text { sup }\left|s_{a}-\sigma\right| \leq M \epsilon$$ Since \(\epsilon\) was arbitrary, it implies that: $$\lim s_{n}=\sigma$$ Thus, we've proven that under the weaker hypothesis, our desired claim holds: if \(\lim \sigma_n=\sigma\) and the provided conditions are satisfied, then \(\lim s_n=\sigma\).

Key concepts

Arithmetic Means

Arithmetic means are a fundamental concept in mathematics. When you calculate an arithmetic mean, you are essentially finding the average of a set of numbers. This process involves adding up all the numbers in the set and then dividing by the number of elements. For example, consider a sequence of numbers, \(\{s_k\}\), and its arithmetic means, \(\sigma_n\), defined as:
\[\sigma_{n}=\frac{s_{0}+s_{1}+\cdots+s_{n}}{n+1}\]
This formula shows that to find the arithmetic mean of the first \(n+1\) terms of a sequence, you add all the terms from \(s_0\) to \(s_n\) together and divide by \(n+1\). This approach is useful in many fields, including finance, physics, and statistics, where averages and means provide insights into the data.

• It helps determine the central tendency of a data set.
• Essential in smoothing out fluctuations within data points.

Understanding arithmetic means is crucial, as it aids in analyzing patterns and predicting future data points in various mathematical sequences.

Complex Sequences

A complex sequence is made up of numbers that have both real and imaginary parts. Each term in a complex sequence, such as \(s_n = a_n + b_ni\), where \(a_n\) and \(b_n\) are real numbers, gives more layers to the analysis than a simple real number.
When we study complex sequences, we often look for patterns and behaviors over time.

• Complex sequences can diverge or converge, similar to real sequences.
• Real-life applications include signal processing and quantum mechanics.

By analyzing complex sequences, we gain deeper insights into phenomena that cannot be captured by real sequences alone. This complexity makes them valuable in mathematical fields and applications where multi-dimensional analysis is required.

Limit of a Sequence

One of the cornerstones in studying sequences is understanding limits. The limit of a sequence refers to the value that the sequence approaches as the index, \(n\), becomes infinitely large. In mathematical terms, for a sequence \(\{s_n\}\), if it converges to a limit \(s\), we write:
\[\lim_{n \to \infty} s_n = s\]
This notation indicates that as \(n\) grows, the sequence values get closer to \(s\). It is essential to establish whether a sequence converges because it can significantly affect the interpretation of the sequence’s behavior over time.In the context of the exercise, establishing that \(\lim \sigma_n = s\) when \(\lim s_n = s\) means that the calculated arithmetic means converge to the same limit as the sequence itself.

• Determining convergence helps predict long-term behavior.
• Applications range from defining series in calculus to economic forecasting.

By mastering the concept of limits, students can better understand and analyze infinite sequences, a critical component in various mathematical and scientific explorations.

Harmonic Sequence

A harmonic sequence is formed by the reciprocals of an arithmetic sequence, usually in the form \(\left\{\frac{1}{n}\right\}_{n=1}^{\infty}\). This type of sequence is known for its divergence; it doesn’t have a finite limit as \(n\) approaches infinity. Despite this divergence, the mean of its partial sums, called the harmonic means, can converge.
Given a sequence \(\{s_n = \frac{1}{n}\}\), the arithmetic means sequence \(\{\sigma_n\}\) becomes:
\[\sigma_n = \frac{1 + \frac{1}{2} + \dots + \frac{1}{n}}{n+1}\]
While the harmonic sequence itself diverges, the arithmetic means \(\sigma_n\) of this sequence may still converge, as analyzed in the exercise where \(\lim \sigma_n = 0\).

• The harmonic sequence is fundamental in number theory and analysis.
• Understanding its divergence is critical for calculating harmonic series.

Through harmonic sequences, students gain insight into contrasting convergence behaviors, contributing to a broader understanding of sequence analysis.