Showing that $\lim_{n\to\infty}\frac{a_1 b_1 +…+a_n b_n}{b_n}=0$ given that $\sum a_n$ converges, $b_n>0$ is monotone increasing, and $\lim b_n=\infty$.
Attempt:
I thought I might as well choose an integer $N$ large so that for $n\geq N$ we have $|a_n| <\epsilon$
$\frac{a_1 b_1 +…+a_n b_n}{b_n}=\frac{\sum_{k=1}^{N} a_k b_k +\sum_{k=N+1}^{n} a_k b_k}{b_n}=\frac{\sum_{k=1}^{N} a_k b_k}{b_n}+\frac{\sum_{k=N+1}^{n} a_k b_k}{b_n}$
Now, for the first term I am not worried at all because the summation is a finite number and I can just make $b_n$ large.
So I am struggling with them term $\frac{\sum_{k=N+1}^{n} a_k b_k}{b_n}$
By Summation by Parts: $\frac{\sum_{k=N+1}^{n} a_k b_k}{b_n}=\frac{\sum_{k=N+1}^{n-1} A_k (b_k-b_{k+1}) + A_n b_n-A_n b_{N+1}}{b_n}$
From here is where I get stuck. I tried expanding out some of the terms but I was not able to sense anything. The summation is begging to use the monotonicity assumption of $\{b_n\}$ but I must be blind.
Thanks for the help!
Best Answer
Let $A_{n}=\displaystyle\sum_{k=1}^{n}a_{k}-\sum a_{k}$, given $\epsilon>0$, there is an $N$ such that $|A_{n}|<\epsilon$ for $n\geq N$.
Now we write \begin{align*} \sum_{k=N}^{n}a_{k}b_{k}&=\sum_{k=N}^{n}(A_{k}-A_{k-1})b_{k}\\ &=A_{N}b_{N}-\sum_{k=N}^{n-1}A_{k}(b_{k+1}-b_{k}), \end{align*} so \begin{align*} \left|\sum_{k=N}^{n}a_{k}b_{k}\right|\leq\epsilon\cdot b_{N}+\epsilon\sum_{k=N}^{n-1}(b_{k+1}-b_{k})=\epsilon\cdot b_{N}+\epsilon\cdot(b_{n}-b_{N}), \end{align*} and also that \begin{align*} \dfrac{1}{b_{n}}\left|\sum_{k=N}^{n}a_{k}b_{k}\right|\leq\epsilon+\epsilon=2\epsilon. \end{align*} But \begin{align*} \left|\dfrac{a_{1}b_{1}+\cdots+a_{n}b_{n}}{b_{n}}\right|\leq\dfrac{1}{b_{n}}\left|\sum_{k=1}^{N}a_{k}b_{k}\right|+\dfrac{1}{b_{n}}\left|\sum_{k=N}^{n}a_{k}b_{k}\right|\leq\dfrac{1}{b_{n}}\left|\sum_{k=1}^{N}a_{k}b_{k}\right|+2\epsilon, \end{align*} so \begin{align*} \limsup_{n\rightarrow\infty}\left|\dfrac{a_{1}b_{1}+\cdots+a_{n}b_{n}}{b_{n}}\right|\leq 2\epsilon. \end{align*} The arbitrariness of $\epsilon>0$ gives \begin{align*} \lim_{n\rightarrow\infty}\left|\dfrac{a_{1}b_{1}+\cdots+a_{n}b_{n}}{b_{n}}\right|=0. \end{align*}