$\lim _{n \rightarrow \infty} \frac{1}{a_{n}} \sum_{k=1}^{n} b_{k}=0$ given existence of limit of sum of ratios Let $\left(a_{n}, n \geq 1\right),\left(b_{n}, n \geq 1\right)$ be sequences of real numbers with $a_{n+1} \geq a_{n}>0$ for every $n \geq 1$ and $\lim _{n \rightarrow \infty} a_{n}=+\infty$. If $\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{b_{k}}{a_{k}}$ exists, then $\lim _{n \rightarrow \infty} \frac{1}{a_{n}} \sum_{k=1}^{n} b_{k}=0$.
I tried using the lemma
$$
\text { if } \lim _{n \rightarrow \infty} v_{n}=v, \quad \text { then } \quad \lim _{n \rightarrow \infty} \frac{1}{a_{n}} \sum_{k=1}^{n} v_{k}\left(a_{k}-a_{k-1}\right)=v
$$
with $v_{n}=\sum_{k=1}^{n} \frac{b_{k}}{a_{k}}$
 A: Let $v = \sum_{k=1}^\infty \frac{b_k}{a_k}$.  Using summation by parts with $v_0 = 0$ and $v_k = \sum_{j=1}^k \frac{b_j}{a_j}$ for $k > 0$, we have
$$\sum_{k=1}^n b_k = \sum_{k=1}^n a_k\frac{b_k}{a_k} =  \sum_{k=1}^n a_k(v_k - v_{k-1}) = a_nv_n - \sum_{k=1}^{n-1}v_k(a_{k+1} - a_{k}),$$
and
$$\frac{1}{a_n}\sum_{k=1}^nb_k = v_n - \frac{1}{a_n}\sum_{k=1}^{n-1}v_k(a_{k+1} - a_{k})$$
Given $\epsilon > 0$ there exists a positive integer $N$ such that for all $k > N$ we have $v - \epsilon \leqslant v_k \leqslant v + \epsilon$. Hence,
$$\frac{1}{a_n}\sum_{k=1}^{n-1}v_k(a_{k+1} - a_{k})=  \frac{1}{a_n}\sum_{k=1}^{N}v_k(a_{k+1} - a_{k}) + \frac{1}{a_n}\sum_{k=N+1}^{n-1}v_k(a_{k+1} - a_{k}) \\\leqslant \frac{1}{a_n}\sum_{k=1}^{N}v_k(a_{k+1} - a_{k}) + (v+ \epsilon) \frac{1}{a_n}\sum_{k=N+1}^{n-1}(a_{k+1} - a_{k}) \\= \frac{1}{a_n}\sum_{k=1}^{N}v_k(a_{k+1} - a_{k}) +(v+\epsilon)\frac{a_n -a_{N+1}}{a_n},$$
and, similarly,
$$\frac{1}{a_n}\sum_{k=1}^{n-1}v_k(a_{k+1} - a_{k}) \geqslant\frac{1}{a_n}\sum_{k=1}^{N}v_k(a_{k+1} - a_{k}) +(v-\epsilon)\frac{a_n -a_{N+1}}{a_n}$$
Since $a_n \to \infty$ as $n \to \infty$, we have, with $N$ fixed ,   $\frac{1}{a_n}\sum_{k=1}^{N}v_k(a_{k+1} - a_{k}) \to 0$, $\frac{a_n -a_{N+1}}{a_n} \to 1$ and
$$v - \epsilon \leqslant \liminf_{n \to \infty}\frac{1}{a_n}\sum_{k=1}^{n-1}v_k(a_{k+1} - a_{k})\leqslant \limsup_{n \to \infty}\frac{1}{a_n}\sum_{k=1}^{n-1}v_k(a_{k+1}-a_k)\leqslant v+\epsilon$$
Since $\epsilon > 0$ can be arbitrarily close to $0$, it follows that
$$\lim_{n \to \infty}\frac{1}{a_n}\sum_{k=1}^{n-1}v_k(a_{k+1} - a_{k})  = v, $$
and
$$\lim_{n \to \infty} \frac{1}{a_n}\sum_{k=1}^nb_k = \lim_{n \to \infty}v_n - \lim_{n \to \infty}\frac{1}{a_n}\sum_{k=1}^{n-1}v_k(a_{k+1} - a_{k}) = v-v = 0$$
