# Let $(s_n)$ be a sequence of nonnegative numbers, and $\sigma_n=\frac{1}{n}(s_1+s_2+\cdots +s_n)$. Show that $\liminf s_n \le \liminf \sigma_n$.

Let $(s_n)$ be a sequence of nonnegative numbers, and for each $n$ define $\sigma_n=\frac{1}{n}(s_1+s_2+\cdots +s_n)$.

Show that $\liminf s_n \le \liminf \sigma_n$.

Actually, I showed $\limsup \sigma_n \le \limsup s_n$ in the following way.

For any $n \gt M \gt N$, we can get the following inequality,

$\sup \{\sigma_n: n\gt M\}\le \frac{1}{M}(s_1+s_2+\cdots +s_N)+\sup\{s_n:n\gt N\}.$

So first taking the limit as $M\to \infty$ then as $N \to \infty$, we get the inequality.

However, this method does not work out for the $\liminf$ case, since the above inequality was derived using the fact that $1/n \lt 1/M$. How can I show the inequality for the $\liminf$ case? I would greatly appreciate any suggestions or solutions.

• – Martin Sleziak Oct 10 '15 at 12:22
• BTW if you already have the proof for limit superior and if use this fact the sequence $(-s_n)$, you get the result for limit inferior. You only need to use this fact. – Martin Sleziak Oct 10 '15 at 12:24

Let $\liminf s_n=s$, $\liminf\sigma_n=\sigma$ and suppose $\sigma<s$. Pick $t\in (\sigma,s)$. Then for infinitely many $n$, $\sigma_n<t$. Pick $l\in(t,s)$. Then for infinitely many $n$, $s_n<l$ and hence $s\leq l<s$. Contradiction.

OR

For any $\epsilon>0$, for $n$ large enough $s-\epsilon< s_n$. Hence for any $\epsilon_1>0$ for $n$ large enough $s-\epsilon-\epsilon_1<\sigma_n$ . Hence $s-\epsilon-\epsilon_1\leq\sigma$. Since this holds for any $\epsilon$,$\epsilon_1$, $s\leq\sigma$.

• Let me allow to explain the proof to amplify the answer. We should notice that if the mean of $x_1$, $\cdots$, $x_n$ are less than $y$ then one of $x_i$ should be less than $y$. – Hanul Jeon Sep 29 '15 at 3:34
• I don't understand the part in the second sentence, where it says for infinitely many $n$, $s_n \lt l$ and hence $s \le l \lt s$. How is this true for any $l \in (t,s)$? – nomadicmathematician Sep 30 '15 at 12:43
• Also in the second solution, why does it follow that for any $\epsilon_1 \gt 0$ for $n$ large enough $s-\epsilon -\epsilon_1 \lt \sigma_n$? Sorry, I'm not really following your answers. I also posted my renewed attempt, could you check if it's correct? – nomadicmathematician Sep 30 '15 at 12:52

I've thought of a solution similar to the one used for proving the limsup case, however, I'm not sure if this is correct.

For any $n\gt N$, $\sigma_n=\frac{s_1+s_2+\cdots s_N+ \cdots s_n}{n}\ge \frac{s_1+\cdots +s_N}{n}+\frac{n-N}{n}\dot \inf\{s_n:n\gt N\}$.

Hence we have $\inf\{\sigma_n: n\gt N\} \ge \frac{s_1+\cdots +s_N}{n}+\frac{n-N}{n}\dot \inf\{s_n:n\gt N\}$.

Now letting $n\to \infty$, we have $\inf\{\sigma_n: n\gt N\} \ge \inf\{s_n:n\gt N\}$. And then letting $N\to \infty$, we get the desired inequality.

• The second and third inequalities do not hold. – A.S. Sep 30 '15 at 13:38
• @A.S. why do they not hold? – nomadicmathematician Oct 3 '15 at 8:22
• Because going from the first inequality to the second you need to take $\inf$ of both sides - not just of the left as you did. – A.S. Oct 3 '15 at 9:59
• I could, but your method works too. Just apply $\inf$ to both sides of first inequality and on the right side use $\inf(A+B)\geq \inf A+\inf B$. Proceed as you already did. – A.S. Oct 3 '15 at 10:55