0
$\begingroup$

Assume that $a_n \ge 0 $, $\{\lambda_n\ge 0\}$ are sequences of real numbers such that, $$\sum\limits_{k=0}^{\infty}\lambda_k = \infty$$ and $\prod \limits_{n=1}^\infty a_n \neq $ converges where $a_n \ge 0 $. Then, show that, $$\lim_{n\to\infty} \frac{1}{\sum_\limits{k=0}^{n}\lambda_k} \sum_\limits{k=0}^{n}\lambda_k a_k=1$$

Can anyone give me some hint or an elegant answer?

$\endgroup$

closed as off-topic by Jack, Xam, kingW3, Carl Mummert, user223391 Nov 12 '17 at 13:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jack, Xam, kingW3, Carl Mummert, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ And it still follows from Cesàro-Stolz also in the mentioned form. $\endgroup$ – Jack D'Aurizio Nov 11 '17 at 16:47
  • 1
    $\begingroup$ Precisely, CS implies that the given limit equals $\lim_{n\to +\infty}a_n$ and the convergence of $\prod a_n$ implies that $\lim_{n\to +\infty}a_n=1$. $\endgroup$ – Jack D'Aurizio Nov 11 '17 at 16:49
  • $\begingroup$ Also, please put all the assumptions ($a_n\geq 0,\lambda_n\geq 0, \prod_{n\geq 1}a_n\text{ is convergent },\sum_{k\geq 0}\lambda_k\text{ is divergent}$) in one place and the claim somewhere else, possibly with some actual attempt from your side. $\endgroup$ – Jack D'Aurizio Nov 11 '17 at 17:32
  • $\begingroup$ Can we have some clarity as to what definition of convergent product is used? proofwiki.org/wiki/… $\endgroup$ – rtybase Nov 12 '17 at 9:29
  • $\begingroup$ I know that the infinite product of positive rwal number converges only when its limits is different from zero. and it written in your link. $\endgroup$ – Guy Fsone Nov 12 '17 at 9:41
1
$\begingroup$

Well, let $p_n=\prod_{i=1}^na_i$. Since $(p_n)_{n\in\mathbb{N}}$ is convergent, it is also a Cauchy sequence. Also, let $$s_p:=\inf_{n\in\mathbb{N}}p_n$$

It is clear that $s_p>0$; otherwise, since $p_n>0$ - a product is convergent if the corresponding sequence converges to a non-zero real number - if $s_p=0$, ther should be a subsequence $(p_{k_n})_{n\in\mathbb{N}}$ of $p_n$, such that $p_{k_n}\to s_p=0$ and, since $p_n$ is convergent, we should have $p_n\to0$, which is a contradiction. So, we get that $s_p>0$.

Now, let $\epsilon>0$; then there exists a $n_0=n_0(\epsilon)\in\mathbb{N}$ such that for every $n,m\in\mathbb{N}$ with $n>m\geq n_0$: $$|p_n-p_m|<\epsilon\cdot s_p$$ or, in other words (taking into consideration that $a_n\geq0$ and that a product converges when its limit is non-zero (so every $a_n$ is strictly positive): $$\left|\prod_{i=1}^na_i-\prod_{j=1}^ma_j\right|<\epsilon\cdot s_p$$ We can re-write the left part of the above inequality as follows: $$\left|\prod_{i=1}^na_i-\prod_{j=1}^ma_j\right|=\left|\prod_{i=1}^ma_i\left(\prod_{j=m+1}^na_j-1\right)\right|=\prod_{i=1}^ma_i\left|\prod_{j=m+1}^na_j-1\right|$$ So, we get the following inequality: $$p_m\left|\prod_{j=m+1}^na_j-1\right|<\epsilon\cdot s_p$$ and, sine all the above numbers are positive and $p_m\geq s_p$, we get that: $$\left|\prod_{j=m+1}^na_j-1\right|<\epsilon\tag{$\star$}$$ Choosing $n=m+1$, $(\star)$ gives: $$|a_{m+1}-1|<\epsilon$$ for every $m\geq n_0\Rightarrow m+1\geq n_0+1$. So, let $N=n_0+1$. Then, for every $n\geq N$ we have: $$|a_n-1|<\epsilon$$ So, we finally get that: $$\boxed{\lim_{n\to\infty}a_n=1}$$

Now, to prove the congergence of $$\frac{\displaystyle\sum_{k=0}^\infty\lambda_ka_k}{\displaystyle\sum_{k=0}^\infty\lambda_k}$$ to $1$ - which, intuitively is obvious, since we are examining the convergence of some convex combinations of a sequence that is convergent to 1 - we can work as follows. At first, let us assume that $\lambda_0>0$ ($\sum_k\lambda_k=\infty$ implies that we can assume that there exists at least one non-zero $\lambda_k$, and, since we need to define the above fraction of sums, we need the denominator to be non-zero for every $n=0,1,\dots$). Also, let: $$s_n:=\frac{\displaystyle\sum_{k=0}^n\lambda_k a_k}{\displaystyle\sum_{k=0}^n\lambda_k}=\sum_{k01}^n\frac{\lambda_k}{\sum_{k=0}^n\lambda_k}a_k=\sum_{k=0}^nc_{k,n}a_k$$ and note the for every $k,n\in\mathbb{N}$ we have that $0<c_{k,n}\leq1$ and $\displaystyle\sum_{k=0}^nc_{k,n}=1$.

Now, let $\epsilon>0$. There exist a $n_0\in\mathbb{N}$ such that for every $n\in\mathbb{N}$ with $n\geq n_0$ we have: $$|a_n-1|<\epsilon\Leftrightarrow1-\epsilon<a_n<1+\epsilon$$ Since $c_{k,n}>0$ for every $k\in\mathbb{N}$ and $n\geq n_0$, we get: $$(1-\epsilon)c_{k,n}<c_{k,n}a_k<(1+\epsilon)c_{k,n}$$ Summing now over $k=0,1,\dots,n$, with $n\geq n_0$, we get: $$\sum_{k=0}^n(1-\epsilon)c_{k,n}<\sum_{k=0}^nc_{k,n}a_k<\sum_{k=0}^n(1+\epsilon)c_{k,n}$$ or $$(1-\epsilon)\sum_{k=0}^nc_{k,n}<s_n<(1+\epsilon)\sum_{k=0}^nc_{k,n}$$ Since $\displaystyle\sum_{k=0}^nc_{k,n}=1$, we get: $$1-\epsilon<s_n<1+\epsilon\Leftrightarrow|s_n-1|<\epsilon$$ for every $n\geq n_0$, and, as a result: $$\boxed{\lim_{n\to\infty}s_n=1}$$ which is what we wanted to prove.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.