This is from baby Rudin. For $\{s_n\}$ a complex sequence , we define the arithmetic means by $\sigma_n$.

$\sigma_n= (s_0+s_1+s_2+...+s_n)/(n+1)$, $n=0,1,2,\ldots$

I am trying to show $\lim s_n= s$ implies $\lim \sigma_n =s$

I thought the arithmetic mean would be between $\limsup s_n$ and $\liminf s_n$ and since the limit exists for the sequence $s_n$ we could deduce that $\lim \sigma_n$ would be sandwiched between those two and as $n \to \infty$, $\lim \sigma_n = s$

Is my reasoning correct? I am not sure if I can say that the arithmetic mean should be between $\limsup$ and $\liminf$ without a proof (is there an easy proof to show that?). Thank you.

  • 1
    $\begingroup$ Look at the Stolz-Cesàro Theorem. $\endgroup$ – robjohn Nov 13 '17 at 5:15
  • $\begingroup$ Where can I find a definition of $\limsup$ & $\liminf$ for a complex sequence? $\endgroup$ – CiaPan Nov 13 '17 at 8:31

Consider first a real sequence $(s_n)_n$ converging to $s.$ For $r>0$ let $N_r\in \Bbb N$ such that $\forall n>N_r\;(s_n\in [-r+s,r+s]).$

For brevity let $A_{n,r}=\frac {1}{1+n}\sum_{j=0}^{N_r}s_j$ and let $T(n)=\frac {1}{1+n}\sum_{j=0}^ns_j$.Then for $n>N_r$ we have $$ A_{n,r}+\frac {n-N_r}{1+n}(s-r)\leq T(n)\leq A_{n,r} +\frac {n-N_r}{1+n}(s+r).$$ Letting $n\to \infty$ (with $r$ and $N_r$ fixed) we have $A_{n,r}\to 0$ and $\frac {n-N_r}{1+n}\to 1$ so $$s-r\leq \lim \inf T(n)\leq \lim \sup T(n)\leq s+r.$$ Since this holds for all $r>0$ we have $\lim_{n\to \infty}T(n)=s.$

For a complex sequence $(c_n)_n$ converging to $c=s+it$ with $s,t\in \Bbb R,$ let $c_n=s_n +it_n$ with $s_n,t_n \in \Bbb R.$ Then $s_n\to s$ and $t_n\to t.$ So by the above result, $(n+1)^{-1}\sum_{j=0}^nc_j=$ $=(n+1)^{-1}\sum_{j=0}^ns_n+i(n+1)^{-1}\sum_{j=0}^nt_n$ converges to $s+it.$

  • $\begingroup$ BTW . If $(s_n)_n$ is $any$ bounded sequence, let $s^+=\lim \sup s_n$ and $s^-=\lim \inf s_n. $ Take $N_r$ such that $\forall n>N_r (-r+s^-\leq s_n\leq r+s^+).$...... In the 3rd line of my A replace the left-hand "$s$" with $s^-$ and replace the right-hand "$s$" with $s^+.$..... Then we conclude that $s^-\leq \lim \inf T(n)\leq \lim \sup T(n)\leq s^+.$ $\endgroup$ – DanielWainfleet Nov 13 '17 at 6:44
  • $\begingroup$ Could you define what $[-r+s, r+s]$ is in the first line for complex $s$...? $\endgroup$ – CiaPan Nov 13 '17 at 9:46
  • $\begingroup$ @CiaPan I have added the missing first sentence and missing last paragraph that I meant to include, but forgot. $\endgroup$ – DanielWainfleet Nov 13 '17 at 14:23


Your statement and intuition are correct (provided we interpret "arithmetic mean" as $\lim_{n\to\infty}\sigma_n)$ , but your statement is only as intuitively true as the theorem in my opinion.

Let $l = \limsup s_n.$ Then there is an $N$ such that $s_n<l$ for all $n>N.$ Then we have $$\sigma_n < \frac{s_0+\ldots+s_N}{n+1} +\frac{n-N+1}{n+1}l$$ for all $n>N.$


Let $$c_n = s_n - s$$ $$d_n = \sigma_n - s$$

The convergence of $(s_n)$ to $s$ means by definition that for each (arbitrarily small) real $\epsilon$ there exists such (big enough) $N$, that $\forall(n>N) |s_n - s| = |c_n| < \epsilon$.

Similary, the convergence of $(\sigma_n)$ to $s$ is equivalent to: for each $\kappa\in\mathbb R$ there exists such $K$, that $\forall(n>K) |\sigma_n-s|= |d_n| < \kappa$.

We need to prove the latter based on the former, right?

We have $$\sigma_n = \frac{\sum_{i=0}^n s_i}{n+1} = \frac{\sum_{i=0}^n (s + c_i)}{n+1} = s+\frac{\sum_{i=0}^n c_i}{n+1}$$ and $$d_n=\frac{\sum_{i=0}^n c_i}{n+1}.$$

Let's choose some $\kappa>0$ and see if there exists such $\sigma_n$, which deviates from $s$ by less than $\kappa$.
This is equivalent to $|d_n|<\kappa$.

Assume $\epsilon = \kappa/2$. Then there exists some $N$ such that $$\forall(n>N) |c_n| < \epsilon = \kappa/2.$$

We take $\sigma_N$ and define $k=\left\lceil|d_N|/\kappa\right\rceil$. We note that for all $n>2k(N+1)$ $$d_n = \frac{\sum_{i=0}^N c_i + \sum_{i=N+1}^n c_i}{n+1} = \frac{N+1}{n+1}\cdot d_N + \frac 1{n+1}\sum_{i=N+1}^n c_i \tag{*}$$ We know that the modulus of complex numbers is a norm in $\mathbb C$ and it satisfies the triangle inequality. If follows that for two terms: $$|A+B|\le |A| + |B|$$ and by induction for longer sums: $$\left|\sum\limits_i c_i\right|\le \sum\limits_i \left|c_i\right|$$ so we can infer from (*): $$d_n < \frac {N+1}{2k(N+1)}d_N + \frac{n-N}{n+1}\frac\kappa 2 \\ < \frac {d_N}{2k} + \frac\kappa 2 \\ \le \frac\kappa 2 + \frac\kappa 2 \\ =\kappa$$

This is equivalent to $|\sigma_n-s| < \kappa$, which is a defining condition of $(\sigma_n)$ sequence convergence, hence $$\lim_{n\to\infty}\sigma_n = s$$ Q.E.D.


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.