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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?
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.
