# Prove that $\lim\limits_{n \to \infty}\frac{a_1b_1+\cdots+a_nb_n}{b_1+b_2+\cdots+b_n}=a$

Suppose $$(a_n)$$ and $$(b_n)$$ are two given sequences for which $$b_n>0$$ and $$\lim\limits_{n→∞}(b_1+\cdots+b_n)=\infty$$ and $$\lim\limits_{n\to\infty}a_n=a$$. Prove that$$\lim_{n\to\infty}\frac{a_1b_1+\cdots+a_nb_n}{b_1+\cdots+b_n}=a.$$

try:

let $$\epsilon > 0$$

Let $$B_n = \sum_{i=1}^n b_i$$ and given it is that for any $$\alpha > 0$$ we can choose $$N$$ so that $$n > N$$ implies $$B_n > \alpha$$. And since $$a_n \to a$$, take $$M > 0$$ such that for all $$n > M$$ one has $$|a_n - a| < \epsilon/b_n$$.

$${\bf one \; can \; take \; \alpha = n }$$ in the first line:

Now, note that

$$| \dfrac{ a_1 b_1 + ... +a_n b_n }{b_1+... b_n } - a | = | \dfrac{ a_1 b_1 + ... + a_n b_n - a B_n }{B_n} | = \dfrac{1}{B_n} |(b_1 (a_1 - a ) + ... + b_n (a_n -a ) | < \dfrac{ n \epsilon }{ n } = \epsilon$$

for any $$n > \max(N,M)$$

Is this a correct proof?

• Where is this question from? May 8, 2020 at 5:42
• You may not find such $M$ such that $\mid a_n -a \mid < \epsilon/b_n$. May 8, 2020 at 6:26
• Yes. For a single $n>M$ such $\epsilon$ can exist, but not for all $n>M$. Then $\epsilon$ shall vary with $n$. May 8, 2020 at 6:40
• You could also use Stolc-Cesaro theorem. BTW if the main question is to check your proof, you can add (solution-verification) tag to mark this. May 8, 2020 at 7:22
• Martin, that very nice link, thanks! I just wanted to see whether I was working correctly and in general feedback for my solution and how to improve... I certainly dont look for solutions but for understanding May 8, 2020 at 7:32

Almost correct. You may not find such $$M$$ for $$\mid a_n -a \mid < \epsilon /b_n$$ since the LHS depends on $$n$$ also. But you just need $$\mid a_n - a \mid < \epsilon$$ and do the almost same thing. You choose $$M$$ first then you choose $$N$$ such that $$B_n > \dfrac{|(b_1 (a_1 - a ) + ... + b_K (a_K -a ) |}{\epsilon}$$
Now let $$K=\max \{M,N\}$$ then for any $$n>K$$
$$| \dfrac{ a_1 b_1 + ... +a_n b_n }{b_1+... b_n } - a | = | \dfrac{ a_1 b_1 + ... + a_n b_n - a B_n }{B_n} | = \dfrac{1}{B_n} |(b_1 (a_1 - a ) + ... + b_n (a_n -a ) | \leq \dfrac{1}{B_n} |(b_1 (a_1 - a ) + ... + b_K (a_K -a ) |+\dfrac{1}{B_n} |(b_{K+1} (a_{K+1} - a ) + ... + b_n (a_n -a ) |< \epsilon + \epsilon\dfrac{b_{K+1}+\cdots+b_n}{B_n}<2\epsilon$$.