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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.

Arctic Char
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Jacob.Lee
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1 Answers1

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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

$$\lim_{n \to \infty}\sum_{j=1}^n a_j b_{n+1-j} = \lim_{n \to \infty} X_n + \lim_{n \to \infty} Y_n = aS$$

RRL
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