Limit of the arithmetic mean 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.
 A: HINT
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 $n>N.$
A: 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 $s+it.$
A: 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.
