How to prove this limit $\lim_{n\to\infty}\frac{p_1a_n+p_2a_{n-1}+\cdots+p_na_1}{p_1+p_2+\cdots+p_n}=a$

Question

If $p_k>0\quad(k=1,2,\cdots)$ and $$\lim_{n\to \infty}\frac{p_n}{p_1+p_2+\cdots+p_n}=0,\lim_{n\to\infty}a_n=a$$how to prove the limit$$\lim_{n\to\infty}\frac{p_1a_n+p_2a_{n-1}+\cdots+p_na_1}{p_1+p_2+\cdots+p_n}=a$$

Answer 1

Proof. $\blacktriangleleft$ First a preparation.

Toeplitz Theorem. Suppose $n,k \in \mathbb N^*$ and $t_{n,k} \geqslant 0$ for all $n$ and $k \leqslant n$ satisfying that $\sum_{k=1}^n t_{n,k} = 1$ and $\lim_{n} t_{n,k}=0$ for all $k \leqslant n$. If $\lim a_n = a$, then we have $$ x _n = \sum_{k =1}^n t_{n,k} a_k\to a \quad [n \to \infty]. $$

To prove the theorem, it suffices to prove for $a = 0$, otherwise use $a_n - a$ to replace $a$, and the result remain unaltered since $\sum_{k =1}^n t_{n,k} =1$.

Since $a_k \to 0 [k \to \infty]$, there is some $M > 0$ s.t. $|a_k| \leqslant M$. For every $N \leqslant n-2$, we have \begin{align*} \varlimsup_n \left| \sum_{k=1}^n t_{n,k}a_k\right| &\leqslant \varlimsup_n \left|\left(\sum_{k=1}^N + \sum_{k=N+1}^n t_{n,k}\right)a_k\right| \\ &\leqslant M \varlimsup_n \sum_{k=1}^n t_{n,k} + \varlimsup_n \sup_{k \geqslant N} |a_k| \sum_{k = N+1}^n t_{n,k}\\ &\leqslant MN \cdot 0 + \sup_{k\geqslant N}|a_k| \varlimsup_n \sum_{k=1}^n t_{n,k} \quad [\text{because }t_{n,k}\geqslant 0]\\ &= \sup_{k \geqslant N} |a_k| \to 0 \quad [N \to \infty], \end{align*} hence the theorem.

Now back to the question. Let $$ t_{n,k} = \frac {p_{n-k+1}}{p_1 + p_2 + \cdots + p_n}. $$ SInce $p_k > 0$ for all $k$, we have $$ 0 \leqslant t_{n,k}\leqslant \frac {p_{n-k+1}}{p_1 + p_2 + \cdots + p_{n-k+1}} \to 0 \quad [n \to \infty], $$ hence $\lim_n t_{n,k} = 0$ for all $k \leqslant n$ by squeezing theorem. Now the conclusion follows the Toeplitz theorem, since clearly $\sum_{k = 1}^n t_{n,k} = 1$. $\blacktriangleright$

Note: the theorem above could be proved by the classic $\varepsilon$-$N$ definition, of course.

Answer 2

Let $\epsilon >0$. Then, there exists some $N$ such that, for all $n >N$ we have $$|a_n-a| <\epsilon$$

Now, by the second condition, since $N$ is fixed you have $$\lim_n \frac{(a_1-a)p_n}{p_1+...+p_n}+....+ \frac{(a_{N-1}-a)p_{n-N}}{p_1+...+p_n}=0$$

Therefore, there exists some $M$ such that, for all $n>M$ we have $$\left|\frac{(a_1-a)p_n}{p_1+...+p_n}+....+ \frac{(a_{N-1}-a)p_{n-N}}{p_1+...+p_n}\right|< \epsilon$$

Then, for all $n>\max \{ N, M\}$ we have $$\left| \frac{p_1a_n+p_2a_{n-1}+\cdots+p_na_1}{p_1+p_2+\cdots+p_n}-a \right| \leq \\ \frac{p_1}{p_1+...+p_n}|a_n-a|+\frac{p_2}{p_1+...+p_n}|a_{n-1}-a|+..+\frac{p_{n-N+1}}{p_1+...+p_n}|a_N-a|+\\ +\left|\frac{(a_1-a)p_n}{p_1+...+p_n}+....+ \frac{(a_{N-1}-a)p_{n-N}}{p_1+...+p_n}\right|\\ \leq \frac{p_1+....+p_{n-N+1}}{p_1+...+p_n} \epsilon+\epsilon < 2 \epsilon $$