# Prob. 14, Chap. 3, in Baby Rudin: The arithmetic mean of a complex sequence

Here's Prob. 14, Chap. 3, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

If $$\left\{ s_n \right\}$$ is a complex sequence, define its arithmetic mean $$\sigma_n$$ by $$\sigma_n = \frac{s_0 + s_1 + \cdots + s_n}{n+1} \ \ \ (n = 0, 1, 2, \ldots).$$

(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 aditional assumption that $$n a_n \to 0$$. ]

(e) Derive the last condition from a weaker hypothesis: Assume $$M < \infty$$, $$\left\vert n a_n \right\vert \leq M$$ for all $$n$$, and $$\lim \ \sigma_n = \sigma$$. Prove that $$\lim s_n = \sigma$$, by completing the following outline:

If $$m < n$$, then $$s_n - \sigma_n = \frac{m+1}{n-m} \left( \sigma_n - \sigma_m \right) + \frac{1}{n-m} \sum_{i=m+1}^n \left( s_n - s_i \right).$$ For these $$i$$, $$\left\vert s_n - s_i \right\vert \leq \frac{(n-i)M}{i+1} \leq \frac{(n-m-1)M}{m+2}.$$ Fix $$\varepsilon > 0$$ and associate with each $$n$$ the integer $$m$$ that satisfies $$m \leq \frac{n-\varepsilon}{1+ \varepsilon} < m +1.$$ Then $$\frac{m+1}{n-m} \leq \frac{1}{\varepsilon}$$ and $$\left\vert s_n - s_i \right\vert < M \varepsilon$$. Hence $$\lim_{n\to\infty} \sup \left\vert s_n - \sigma \right\vert \leq M \varepsilon.$$ Since $$\varepsilon$$ was arbitrary, $$\lim s_n = \sigma$$.

My effort:

Part (a):

If $$\lim s_n = s$$, then we can find a natural number $$N$$ such that $$n > N$$ implies that $$\left\vert s_n - s \right\vert < 1.$$ So, for $$n > N$$, we have \begin{align} \left\vert \sigma_n - s \right\vert &= \left\vert \frac{ s_0 + s_1 + \cdots + s_n}{n +1} - s \right\vert \\ &\leq \frac{1}{n+1} \left( \left\vert s_0 - s \right\vert + \cdots + \left\vert s_N - s \right\vert + \cdots + \left\vert s_n - s \right\vert \right) \\ &\leq \frac{1}{n+1} \left( (N+1) \max \left\{ \left\vert s_0 - s \right\vert, \ldots, \left\vert s_N - s \right\vert \right\} + \left\vert s_{N+1} - s \right\vert + \cdots + \left\vert s_n - s \right\vert \right) \\ &< \frac{1}{n+1} \left( (N+1) \max \left\{ \left\vert s_0 - s \right\vert, \ldots, \left\vert s_N - s \right\vert \right\} + (n-N) \right) \end{align} Let $$\varepsilon > 0$$ be given. What next?

Part (b):

Let $$s_n = (-1){n+1}$$ for $$n = 0, 1, 2, \ldots$$. Then $$\sigma_n = \begin{cases} \frac{1}{n+1} \ \mbox{ if } n \mbox{ is even}; \\ 0 \ \mbox{ if } n \mbox{ is odd}. \end{cases}$$ Then $$\left\{ s_n \right\}$$ fails to converge, but $$\lim \ \sigma_n = 0$$. Is this example correct?

Part (c):

My feeling is the answer is no, but I cannot establish this rigorously. How to?

Part (d):

If we put $$a_n = s_n - s_{n-1}$$, for $$n \geq 1$$, then $$s_n = a_0 + s_1 + \cdots + a_n$$, and so \begin{align} s_n - \sigma_n &= s_n - \frac{ s_0 + s_1 + \cdots + s_n}{n+1} \\ &= \frac{ (n+1) s_n - s_0 - s_1 - \cdots - s_n }{n+1} \\ &= \frac{ (n+1) \left( s_0 + a_1 + \cdots + a_n \right) - s_0 - \left( s_0 + a_1 \right) - \left( s_0 + a_1 + a_2 \right) - \cdot - \left( s_0 + a_1 + \cdots + a_n \right) }{n+1} \\ &= \frac{ a_1 + 2 a_2 + \cdots + n a_n }{n+1}. \end{align} Now we assume that $$\lim_{n \to \infty} n a_n = 0$$ and that $$\left\{ \sigma_n \right\}$$ converges. How to show that $$\left\{ s_n \right\}$$ convreges?

Part (e):

If $$m < n$$, then \begin{align} & \frac{ m+1 }{ n-m } \left( \sigma_n - \sigma_m \right) + \frac{1}{n-m} \sum_{i = m+1}^n \left( s_n - s_i \right) \\ &= \frac{ m+1 }{ n-m } \left( \frac{ s_0 + s_1 + \cdots + s_n}{n+1} - \frac{ s_0 + s_1 + \cdots + s_m}{m+1} \right) + \frac{1}{n-m} \sum_{i = m+1}^n \left( s_n - s_i \right) \\ &= s_n + \frac{1}{n-m} \left[ (m+1) \frac{ s_0 + \cdots + s_n}{n+1} - \left( s_0 + \cdots + s_n \right) \right] \\ &= s_n - \sigma_n. \end{align} How to proceed from here?

• Part (a) is answered here math.stackexchange.com/questions/565288/… Commented Dec 2, 2016 at 22:54
• For $d,$ consider $\left|\frac{ a_1 + 2 a_2 + \cdots + n a_n }{n+1}\right |\le \frac{n}{n+1}\max\{n|a_1|,\cdots, n|a_n|\}$. Commented Dec 3, 2023 at 0:35

For (c): Consider $s_{2^m} = m, m = 1,2, \dots, s_n = 0$ for all other $n.$ (We don't have $s_n >0$ for all $n,$ but it will give you the idea.)