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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?

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  • $\begingroup$ Where is this question from? $\endgroup$
    – jimjim
    May 8, 2020 at 5:42
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    $\begingroup$ You may not find such $M$ such that $\mid a_n -a \mid < \epsilon/b_n$. $\endgroup$
    – N.Quy
    May 8, 2020 at 6:26
  • $\begingroup$ Yes. For a single $n>M$ such $\epsilon$ can exist, but not for all $n>M$. Then $\epsilon$ shall vary with $n$. $\endgroup$
    – Alapan Das
    May 8, 2020 at 6:40
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    $\begingroup$ 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. $\endgroup$ May 8, 2020 at 7:22
  • $\begingroup$ 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 $\endgroup$ May 8, 2020 at 7:32

1 Answer 1

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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 $n>K$

$| \dfrac{ a_1 b_1 + ... +a_n b_n }{b_1+... b_n } - a | = | \dfrac{ a_1 b_1 + ... + a_n b_n - a B_n }{B_n} | = \dfrac{1}{B_n} |(b_1 (a_1 - a ) + ... + b_n (a_n -a ) | \leq \dfrac{1}{B_n} |(b_1 (a_1 - a ) + ... + b_K (a_K -a ) |+\dfrac{1}{B_n} |(b_{K+1} (a_{K+1} - a ) + ... + b_n (a_n -a ) |< \epsilon + \epsilon\dfrac{b_{K+1}+\cdots+b_n}{B_n}<2\epsilon$.

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  • $\begingroup$ perfect1 thanks a lot ! $\endgroup$ May 8, 2020 at 6:44
  • $\begingroup$ You’re welcome. ^^ $\endgroup$
    – N.Quy
    May 8, 2020 at 6:45

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