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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?

Thanks in advance.

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  • $\begingroup$ $a_n \to \infty$ 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. $\endgroup$
    – Martin R
    Mar 11 at 12:02
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Your mistake is here:

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

Actually, $$ \lim_{n\to\infty}\frac{a_{n}}{b_{n}}=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac 1 k=0.$$

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    $\begingroup$ So the proof in the last comment here is wrong? math.stackexchange.com/questions/1836226/… $\endgroup$
    – FreeZe
    Mar 11 at 12:01
  • $\begingroup$ @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 $0$. $\endgroup$ Mar 11 at 12:05
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    $\begingroup$ @Kavu Rama Murthty I see. Thanks that is very helpful. $\endgroup$
    – FreeZe
    Mar 11 at 12:07

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