Exercise 3.14

Question

Exercise 14:
    If $\{s_n\}$ is complex sequence, define its arithmetic means $\sigma_n$ by $$\sigma_n=\frac{s_0+s_1+\cdots+s_n}{n+1}\quad\quad(n=0,1,2,\ldots).$$
    (a) If $\lim s_n=s$, prove that $\lim\sigma_n=s$.

(b) Construct a sequence $\{s_n\}$ which does not converge, although $\lim\sigma_n=0$.

(c) Can it happen that $s_n>0$ for all $n$ and that $\limsup s_n=\infty$, although $\lim\sigma_n=0$?

(d) Put $a_n=s_n-s_{n-1}$, for $n\ge1$. Show that $$s_n-\sigma_n=\frac{1}{n+1}\sum_{k=1}^nka_k.$$ Assume that $\lim(na_n)=0$ and $\{\sigma_n\}$ converges. Prove that $\{s_n\}$ converges.

(e) Derive the last conclusion from a weaker hypothesis: Assume $M<\infty$, $|na_n|\le M$ for all $n$, and $\lim\sigma_n=\sigma$. Prove that $\lim s_n=\sigma$.

ghostofgarborg · Accepted Answer

\def\where{\,|\,} \def\eps{\varepsilon} (a) Choose $\epsilon > 0$ and $N$ such that $|s_n - s_m| < \epsilon/2$ for all $n, m > N$. Then choose $M$ such that $$ \left| \frac{(s_0 - s) + \cdots + (s_{N-1} - s)}{M} \right| < \frac \epsilon 2 $$ Then whenever $n > \max(M, N)$ $$ \left| \frac{(s_0 - s) + \cdots + (s_n-s)}{n+1} \right| \leq \left| \frac{(s_0 - s) + \cdots + (s_{N-1} - s)}{n+1} \right| + \left| \frac{(S_N-s) + \cdots + (S_n-s)}{n+1} \right| < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon $$ This shows that $(\sigma_n -s)$ converges to $0$, i.e. that $\sigma_n \to s$. (b) Let $s_n = (-1)^n$. Then $s_n$ does not converge, but $\sigma_n = \frac{1 + (-1)^{n}}{2(n+1)}$ converges to $0$. (c) Yes. Let $a_n = \frac 1 {2^n}$, $b_{2^n} = n$, and $ b_n = 0$ when $n$ is not a power of $2$. Let $s_n = a_n + b_n$. Then $s_n > 0$, and the series is unbounded, so $\limsup s_n = \infty$. However $$ 0 < \sigma_n = \frac 1 {n+1} \sum_{i=0}^n \frac {1}{2^i} + \frac{1}{n+1} \sum_{i=0}^{\lfloor \log_2(n) \rfloor} i < \frac{2}{n} + \frac{\log_2(n)^2}{n} $$ where the right hand side tends to 0. This shows that $\sigma_n \to 0$. (d) Note that $$ \sum_{k=1}^n k(s_k - s_{k-1}) \ = \ \sum_{k=1}^n ks_k - \sum_{k=0}^{n-1} (k+1)s_k \ = \ ns_n - \sum_{k=0}^{n-1}s_k \ = \ (n+1)s_n - \sum_{k=0}^n s_k $$ so that $$ s_n - \sigma_n = \frac{1}{n+1} \sum_{k=0}^n k a_k $$ Note that the right hand side is the $n$th arithmetic mean of the sequence $na_n$, which by assumption and our result in (a) converges. Therefore $s_n - \sigma_n$ is convergent, and since $\sigma_n$ is convergent by assumption, $(s_n - \sigma_n) + \sigma_n = s_n$ is convergent. (e) \begin{align*} s_n - \sigma_n &= s_n - \sum_{k=0}^n \frac{s_k}{n+1} \ &= s_n - \sum_{k=0}^n \left(\frac{1}{n-m} - \frac{m+1}{(n+1)(n-m)} \right) s_k \ &= \sum_{k=0}^m \left(\frac{(m+1)s_k}{(n+1)(n-m)} - \frac{s_k}{n-m} \right) + \sum_{k=m+1}^n \left(\frac{(m+1)s_k}{(n+1)(n-m)} + \frac{s_n - s_k}{n-m} \right) \ &= \frac{m+1}{n-m} \left( \sum_{k=0}^n \frac{s_k}{n+1} - \sum_{k=0}^m \frac{s_k}{m+1} \right) + \frac{1}{n-m} \sum_{k=m+1}^n (s_n - s_k) \end{align*} which is the desired equality. By assumption $|s_n - s_{n-1}| < M/n$, so that $$ |s_n - s_t| = | (s_n - s_{n-1}) + (s_{n-1} - s_{n-2}) + \cdots + (s_{i+1} - s_i)| \leq \frac M n + \cdots + \frac {M}{i+1} \leq \frac{(n-i)M}{i+1} $$ The rest of the argument is covered in sufficient detail in the text.

Exercise 3.14

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Exercise 3.14

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