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Let $(a_n)_{n\in \mathbb{N^*}}$ a strictly positive sequence such that $\displaystyle : \lim_{n\to +\infty} \sum _{k=1}^n a_k=+\infty$. Show that for all sequence $(b_n)_{n\in \mathbb{N^*}}$ of real numbers such as $\displaystyle \lim_{n \to +\infty} b_n = l \in \mathbb{\bar{R}}$, we have $:$ $$\lim_{n\to +\infty} \frac{a_1b_1+a_2b_2+a_3b_3+...+a_nb_n}{a_1+a_2+a_3+...+a_n}=l$$

  • I tried to use some inequalities such as the Cauchy inequality which gave me $\sum a_kb_k\le \sqrt{\sum a_k^2\ \cdot \ \sum b_k^2 }$
  • I also tried the definition of the limit with Cesaro lemma yet it doesnt work

Any idea ? thanks

PortoKranto
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2 Answers2

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Let $u_n = \sum_{k=1}^n a_k$, and $v_n = b_{n+1}-b_n$. You have that $(u_n)$ increases to $+\infty$, and $v_n$ is the general term of a convergent series, hence by Kronecker's lemma, you deduce $$\lim_{n \rightarrow +\infty}\frac{\sum_{k=1}^n u_kv_k}{u_n}=0 \quad \quad (*)$$

But $$\sum_{k=1}^n u_kv_k = \sum_{k=1}^n u_k(b_{k+1}-b_k)=u_nb_{n+1} - u_1b_1 + \sum_{k=2}^{n} (u_{k-1}-u_k)b_{k}=u_nb_{n+1} - u_1b_1 - \sum_{k=2}^{n} a_kb_{k}$$

so $$ \frac{\sum_{k=1}^n u_kv_k}{u_n} = b_{n+1} - \frac{u_1b_1}{u_n}-\frac{\sum_{k=2}^{n} a_kb_{k}}{u_n}=b_{n+1} - \frac{u_1b_1}{u_n}-\frac{\sum_{k=2}^{n} a_kb_{k}}{\sum_{k=1}^n a_k}$$

So using $(*)$,$$\lim_{n \rightarrow +\infty} \frac{\sum_{k=2}^{n} a_kb_{k}}{\sum_{k=1}^n a_k} = \lim_{n \rightarrow +\infty} b_{n+1} = l$$

TheSilverDoe
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The proof is very similar to showing that if $b_n \to l$, then $\overline{b_n}=\frac{b_1+\ldots+b_n}{n} \to l.$ In fact, this proof is just a special case of your problem, but with $a_k=1$ for all $k$.

Given $\epsilon>0$, choose $n_0$ sufficiently large so that $|b_n-l|<\frac{\epsilon}{2}$ for all $n> n_0$.

Write $\left|\frac{a_1b_1+a_2b_2+a_3b_3+...+a_nb_n}{a_1+a_2+a_3+...+a_n}-l\right|=\left|\frac{a_1(b_1-l)+a_2(b_2-l)+a_3(b_3-l)+...+a_n(b_n-l)}{a_1+a_2+a_3+...+a_n}\right|\leq A(n) + B(n)$, where \begin{align} A(n)&=\frac{\sum_{k=1}^{n_0}a_k|b_k-l|}{a_1+a_2+a_3+...+a_n}, \\ B(n)&=\frac{\sum_{k=n_0+1}^{n}a_k|b_k-l|}{a_1+a_2+a_3+...+a_n} \end{align} Since $b_k \to l$, see if you can prove that the numerator in $A(n)$ is bounded and the denominator blows up.

For $B(n)$, see if you can prove $B(n)<\frac{\epsilon}{2}$.

ProfOak
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