Krdytkyu

Suppose $p_i >0 ~(i=1,2,ldots)$ and $lim_{n to infty}{frac{p_n}{p_1+p_2+...+p_n}}=0$ and $lim_{n to infty}{a_n}=a$ prove that $$lim_{n to infty}{frac{p_1a_n+p_2a_{n-1}+...+p_n a_1}{p_1+p_2+...+p_n}}=a.$$

Now proof attempt:
$|a_n|$ is bounded as it converges. So:
$$left|frac{p_1a_n+p_2a_{n-1}+...+p_n a_1}{p_1+p_2+...+p_n}-aright|=left|frac{p_1(a_n-a)+p_2(a_{n-1}-a)+...+p_n (a_1-a)}{p_1+p_2+...+p_n}right| leq \
leq frac{|p_1(a_n-a)|+|p_2(a_{n-1}-a)|+...+|p_n (a_1-a)|}{p_1+p_2+...+p_n},$$
and for some $N$ and $M$ we have:
$$frac{|p_1(a_n-a)|+|p_2(a_{n-1}-a)|+...+|p_{n-N} (a_{N+1}-a)|+p_{n-N+1}M+...+p_nM}{p_1+p_2+...+p_n}.$$
So now what do I do?

You have started well and your relation holds true for a fixed natural $N$ and any $n geq N$. Since $N$ is fixed the part with $frac{(p_{n-N+1}+...+p_n)M}{p_1+...+p_n}$ will go to zero by your hypothesis(it is a sum of finitely many terms which each go to zero).

For the second term in your fraction, note that $(a_{N+1}-a),...,(a_n-a)$ are all bounded by some $epsilon$ and that $frac{p_1+...+p_{n-N}}{p_1+...+p_n}$ tends to one so the whole fraction $frac{|p_1(a_n-a)+...+p_{n-N}(a_{N+1}-a)|}{p1+...+p_n}$ is bounded above by $epsilonfrac{p_1+...+p_{n-N}}{p1+...+p_n}$ and goes to zero, whence your proof.