# A question about Stolz-Cesàro theorem

Stolz-Cesàro theorem states that given some sequence $$\left(a_{n}\right)_{n\geq1}$$ and a monotone strictly increasing sequence that diverge $$\left(b_{n}\right)_{n\geq1}$$ , such that $$\lim_{n\to\infty}\frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=L$$

Then the following holds:

$$\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=L$$

Now, given a divergent sum $$\sum_{n=1}^{\infty}a_{n}$$, we have that also $$\frac{1}{n}\sum_{k=1}^{n}a_{k}$$ diverge to $$\infty$$ as we can see here

So consider $$a_{n}=\sum_{k=1}^{n}\frac{1}{k}$$ and $$b_{n}=n$$. Then

$$\frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=\frac{\frac{1}{n+1}}{1}\underset{n\to\infty}{\longrightarrow}0$$

But $$\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}a_{k}$$

Does not converge to $$0$$.

Hows is this possible consider that Stolz-Cesàro theorem is true? What am I missing?

• $a_n \to \infty$ implies $\frac{1}{n}\sum_{k=1}^{n}a_{k} \to \infty$, but the divergence of $\sum_{n=1}^{\infty}a_{n}$ does not imply that. math.stackexchange.com/q/1836226/42969 cannot be applied in your example. Mar 11 at 12:02

But $$\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}a_{k}$$
Actually, $$\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac 1 k=0.$$
• @FreeZe $\frac 1 n \sum_{k \leq n} a_k \to \infty$ but that is not what we have here.We have $\frac 1 n \sum_{k \leq n} \frac 1k$ and this tends to $0$. Mar 11 at 12:05