How to compute the limit by Stolz theorem? As is well known, Stolz–Cesàro theorem is the following:
Let $\displaystyle {(a_{n})_{n\geq 1}}$ and ${\displaystyle (b_{n})_{n\geq 1}}$be two sequences of real numbers. Assume that ${\displaystyle (b_{n})_{n\geq 1}}$ is a strictly monotone and divergent sequence (i.e. strictly increasing and approaching ${\displaystyle +\infty }$ , or strictly decreasing and approaching ${\displaystyle -\infty }$ and the following limit exists:
$${\displaystyle \lim _{n\to \infty }{\frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}=l.\ }$$
Then, the limit
$${\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=l.\ }$$
I want to know that If we have $${\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=l.\ }$$ Can we deduce that
$${\displaystyle \lim _{n\to \infty }{\frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}=l?\ }$$
If so, we can  reslove the following exercise by the above result:
If $a_n\to a, b_n\geq 0, \forall n\in Z^+, \lim_{n\to \infty} b_1+b_2+\cdots+b_n=S,$ then
$$\displaystyle \lim_{n\to \infty}a_1b_n+a_2b_{n-1}+\cdots+a_nb_{1}=aS.$$
Let $S_n= b_1+b_2+\cdots+b_n$ and  $S_0=0$, then $S_n\to S,$ and
$$a_1b_n+a_2b_{n-1}+\cdots+a_nb_{1}=\sum_{i=1}^{n}a_i(S_{n+1-i}-S_{n-i})$$
Hence,
$\lim_{n\to \infty}a_1b_n+a_2b_{n-1}+\cdots+a_nb_{1}$
$\displaystyle= \lim_{n\to \infty} \frac{\sum_{i=1}^{n}a_i(S_{n+1-i}-S_{n-i})}{n-(n-1)} $
$\displaystyle= \lim_{n\to \infty}\frac{\sum_{i=1}^{n}a_iS_{n+1-i}}{n}=aS.$
If not so, I would appreciate anyone who give the right solution and any suggestions.
 A: I don't believe you can use the converse of Stolz-Cesaro because as discussed in the comment,  $n$ is in the denominator and $\frac{n}{n+1} \to 1$ as $n \to \infty$.
For an alternative, write $a_n = a + \epsilon_n$ where we have $\epsilon_n \to 0$ as $n \to \infty$. We then have
$$\sum_{j=1}^n a_j b_{n+1-j} = \underbrace{a\sum_{j=1}^nb_{n+1-j}}_{X_n} + \underbrace{\sum_{j=1}^n\epsilon_jb_{n+1-j}}_{Y_n}$$
Note that
$$\lim_{n \to \infty} X_n = \lim_{n \to \infty}a\sum_{j=1}^nb_{n+1-j} = \lim_{n \to \infty}a\sum_{j=1}^nb_{j} = aS$$
Since $\epsilon_n \to 0$, there exists $N \in \mathbb{N}$ such that $|\epsilon_n| < \epsilon$ for all $n > N$.
Thus, with $n > N$,
$$\tag{*}|Y_n| = \left|\sum_{j=1}^n\epsilon_jb_{n+1-j}\right| \leqslant \left|\sum_{j=1}^N\epsilon_jb_{n+1-j}\right|+\left|\sum_{j=N+1}^n\epsilon_jb_{n+1-j}\right|\\ \leqslant \left|\sum_{j=1}^N\epsilon_jb_{n+1-j}\right|+\sum_{j=N+1}^n|\epsilon_j|\,|b_{n+1-j}|$$
For the second sum on the RHS of (*), since $b_n \geqslant 0$, we have
$$\sum_{j=N+1}^n|\epsilon_j|\,|b_{n+1-j}| = \sum_{j=N+1}^n|\epsilon_j|\,b_{n+1-j} < \epsilon \sum_{j=N+1}^n\,b_{n+1-j} = \epsilon \sum_{j=1}^{n - N}\,b_{j} < \epsilon S$$
Hence,
$$\tag{**}|Y_n| < \left|\sum_{j=1}^N\epsilon_jb_{n+1-j}\right| + \epsilon S$$
Since, $\sum b_n$ converges we have $b_n \to 0$ as $n \to \infty$. Since $N$ is fixed, for the first sum on the RHS (**) we get
$$\lim_{n \to \infty}\left|\sum_{j=1}^N\epsilon_jb_{n+1-j}\right| = 0 $$
Since $\epsilon >0$ can be arbitrarily small, it follows that $Y_n \to 0$ as $n \to \infty$, and
$$\lim_{n \to \infty}\sum_{j=1}^n a_j b_{n+1-j} = \lim_{n \to \infty} X_n + \lim_{n \to \infty} Y_n  = aS$$
