2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.