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Let $a_n$ be a sequence of real non-negative numbers

Define $S_n = fracsum_i=1^n a_in$

Prove that

$$liminf(a_n) leq liminf(S_n) leq limsup(S_n) leq limsup(a_n)$$

I wanted to show that $infa_n:n geq t leq infS_n:ngeq t$ for all $t$

However this doesn't seem to be true

The inequality seems to hold only at n goes to infinity but not every n individually

Can anyone help me?

The statement follows directly by Stolz–Cesàro theorem

$$liminf fracA_n+1-A_nB_n+1-B_n leq liminf fracA_nB_n leq limsup fracA_nB_n leq limsup fracA_n+1-A_nB_n+1-B_n$$

by $A_n=S_n$ and $B_n=n$.

Let $l$ be the $liminf$ of $a_n$. Then for all $varepsilon > 0$, there exists an $N in mathbbN$ such that
$$n ge N implies a_n ge l - varepsilon.$$
Let $M = sum_k=1^N-1 a_k.$ Then,
$$n ge N implies sum_k=1^n a_n = M + sum_k=N^n a_n ge M + (n - N + 1)(l - varepsilon).$$
Therefore, when $n ge N$, we have
$$fracsum_k=1^n a_nn ge fracMn + left(1 - fracN - 1nright)(l - varepsilon) to l - varepsilon.$$
Hence, every subsequence of $fracsum_k=1^n a_nn$ must converge to at least $l - varepsilon$, where $varepsilon > 0$ is arbitrary. Thus, every subsequence must converge to at least $l$, proving the inequality of the limits inferior.