So I found this question:

If $\lim_{n \to \infty} \frac{c_{n+1}}{n \cdot c_n} = a$ then prove that $\lim_{n \to \infty} \sqrt[n+1]{c_{n+1}} - \sqrt[n]{c_n} = \frac{a}{e}$

Now I used the Stolz - Cesaro theorem to prove this by the following method:

Let $a_{n+1} = \sqrt[n+1]{c_{n+1}}$ and $b_{n+1} = n+1$

Thus $\lim_{n \to \infty} \sqrt[n+1]{c_{n+1}} - \sqrt[n]{c_n} = \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{a_n}{b_n} = \frac{\sqrt[n]{c_n}}{n} = \sqrt[n]{\frac{c_n}{n^n}}$

Let $x_n = \frac{c_n}{n^n}$ then using the converse of the geometric mean case of Stolz - Cesaro theorem $(\lim_{n \to \infty} \sqrt[n]y_n = L \implies \lim_{n \to \infty}\frac{y_{n+1}}{y_n} = L$) we get $\sqrt[n]{\frac{c_n}{n^n}} = \frac {c_{n+1} \cdot n^n}{c_n \cdot (n+1)^{n+1}} = \frac {c_{n+1}}{n \cdot c_n} \cdot \frac{1}{(1 + \frac{1}{n})^{n+1}}$

I don't know if the converse of the multiplicative identity of the Stolz - Cesaro theorem can be used or not but I could not find another way to solve this question. Can you also tell me if the converse of the multiplicative identity of the Stolz - Cesaro theorem exists here?

we know $(1 + \frac{1}{n})^n = (1 + \frac{1}{n})^{n+1} = e\;\&\;(\lim_{n \to \infty} (1+\frac{1}{n}) → 1)$ and it is given that $\frac {c_{n+1}}{n \cdot c_n} = a$

So we get that $\lim_{n \to \infty} \sqrt[n+1]{c_{n+1}} - \sqrt[n]{c_n} = \frac{a}{e}$

Obviously the problem (ignoring the usage of the converse of the multiplicative identity of the Stolz - Cesaro theorem) is we have to prove that $\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$ exists to use Stolz - Cesaro theorem but I cannot find a way to prove how to do it, I just can't get where to start to prove that it exists. It intuitively looks like it should exist.

So my question is : can we prove that it exists and if not, then how would we legitimately solve this question?

P.S. : I saw this question but that solution has a different approach and also my doubt is a little different because I want to use the method stated above. To prove that, we had to prove that $\lim_{n \to \infty} \sqrt[n+1]{c_{n+1}} - \sqrt[n]{c_n}$ exists first which I am not able to do.

  • $\begingroup$ I don't know whether there is a general (partial) converse to the Cesaro-Stolz theorem. There is this article: D. M. Batinetu, Problem lui Lalescu si Reciproca Teoremei Cesaro-Stolz, Lucrarile Seminarului Didactica Matematicii, Universitatea Babes-Bolyai. Cluj-Napoca vol 15/2000, pp. 3-8 (in Romanian) I unfortunately have no access to that paper. Perhaps someone in Romania may provide a link to the article. In any case, I think it may involve problems of the form $\lim_n(a_{n+1}-a_n)$ when $\lim_n\frac{a_n}{n}$ exits plus some other ratio condition. $\endgroup$
    – Mittens
    Nov 21 at 2:08
  • $\begingroup$ The key is to note that if $a_n=c_n^{1/n}$ then $a_{n+1}/a_n\to 1$ (prove this) and then one can write the difference $a_{n+1}-a_n$ as $a_n((a_{n+1}/a_n)-1)$ and replace the parentheses thing by $\log (a_{n+1}/a_n)$. This helps a lot in simplifying the expression. $\endgroup$
    – Paramanand Singh
    Nov 27 at 15:16

2 Answers 2


This is a slight simplification of @Mittens's solution, for the sake of my own reference. So I am posting this in community wiki mode. Feel free to update it by yourself!

1. To begin with, let me first present my take of the proof of @Mittens's result:

Theorem. Let $(a_n)$ be a sequence of positive real numbers such that $$ \lim_{n\to\infty} \frac{a_n}{n} = A \in [0, \infty) \qquad \text{and}\qquad \lim_{n\to\infty} \left(\frac{a_{n+1}}{a_n}\right)^n = B \in (0, \infty). $$ Then $\lim_{n\to\infty} (a_{n+1} - a_n) = A \log B$.

Proof. From the assumption, we get

$$ \log a_{n+1} - \log a_n = \frac{\log B + o(1)}{n}. $$

Then by using the relation $e^x = 1 + x + \mathcal{O}(x^2)$ as $x \to 0$, we get

\begin{align*} a_{n+1} - a_n &= \frac{a_n}{n} \cdot n \left[ \exp(\log a_{n+1} - \log a_n) - 1 \right] \\ &= \frac{a_n}{n} \left[ n (\log a_{n+1} - \log a_n) + \mathcal{O}(n^{-1}) \right] \\ &\to A \log B. \end{align*}

2. Now let $(c_n)$ be as in OP, and let $a_n = \sqrt[n]{c_n}$. Applying Stolz–Cesaro theorem,

\begin{align*} \log\left(\frac{a_n}{n}\right) &= \frac{\log c_n - n\log n}{n} \\ &\sim \frac{[\log c_{n+1} - (n+1)\log(n+1)] - [\log c_n - n\log n]}{(n+1) - n} \\ &= \log \left( \frac{c_{n+1}}{n c_n} \right) - (n+1) \log\left(1 + \frac{1}{n}\right) \\ &\to (\log a) - 1 \end{align*}

as $n \to \infty$, hence we get $A = \frac{a}{e}$. Moreover,

\begin{align*} \left( \frac{a_{n+1}}{a_n} \right)^n &= \frac{n}{a_{n+1}} \cdot \frac{c_{n+1}}{n c_n} \to \frac{a}{A}, \end{align*}

hence $B = \frac{a}{A} = e$. Therefore

$$ a_{n+1} - a_n \to A \log B = \frac{a}{e} $$

as required.

  • $\begingroup$ I don't exactly understand the first proof and also, did you apply the converse of the Stolz - Cesaro theorem in the 2nd part? $\endgroup$
    – Ham Lemon
    Nov 21 at 8:53
  • $\begingroup$ @HamLemon, If you can clarify in which step you have trouble understanding, I can provide explanation. But the key observation is that, the assumption on the convergence of $(\frac{a_{n+1}}{a_n})^n$ gives some explicit control on the asymptotic behavior of $n(\frac{a_{n+1}}{a_n}-1)$, which we can take advantage of. Also, for the second question, the logic goes as follows: Let $b_n := \log c_n - n\log n$. Then a direct computation shows that $$\frac{b_{n+1} - b_n}{(n+1) - n} \to A$$ with $A=(\log a)-1$. Hence by Stolz–Cesaro theorem, $$\frac{b_n}{n} \to A$$ as well. $\endgroup$ Nov 21 at 9:02
  • $\begingroup$ For the 1st part, I have a doubt in how you wrote $loga_{n+1} - loga_n = \frac{logB + O(1)}{n}$ to be equivalent to $a_{n+1}-a_n = \frac{a_n}{n} \cdot n[exp(loga_{n+1} - loga_n)-1]$ $\endgroup$
    – Ham Lemon
    Nov 21 at 9:07
  • $\begingroup$ Also in the second proof, didnt you do $log(\frac{a_n}{n}) = \frac{b_n}{n} \sim \frac{b_{n+1} - b_n}{(n+1)-n}$ which is the converse of the stolz - cesaro theorem? $\endgroup$
    – Ham Lemon
    Nov 21 at 9:10
  • $\begingroup$ @HamLemon, For the first part, let $\varepsilon_n=(a_{n+1}/a_n)^n-B$ denote the explicit difference between the sequence and its limit (which, in my original proof, was abbreviated as $o(1)$). Then a simple algebra gives $$\log a_{n+1} - \log a_n=\frac{\log B + \log(1+\varepsilon_n/B)}{n}, $$ and it is clear that $\log(1+\varepsilon_n/B)\to 0$ as $n\to\infty$. So, this term is again $o(1)$. Also, the equality $$a_{n+1} - a_n = \frac{a_n}{n} \cdot n\left[\exp(\log a_{n+1} - \log a_n) - 1\right] $$ is the result of pure algebraic manipulation, irrelevant of the previous argument. $\endgroup$ Nov 21 at 9:25

This position is to present a special partial converse to the Cesàro-Stolz theorem, and which can be used to provide a simple solution to the problem in the OP

Proposition L: Let $(a_n:n\in\mathbb{N})$ be a sequence of positive numbers such that

  1. $\lim_n\frac{a_n}{n}=A>0$,
  2. $\lim_n\Big(\frac{a_{n+1}}{a_n}\Big)^n=B\in\overline{\mathbb{R}}$.

Then $L=\lim_n(a_{n+1}-a_n)$ exists and equals $A\log B$.

These types of results are very well known to Romanian students and have a long tradition since the early 1900's (Look for Lalescu's problem).

Proof of proposition: The proof of the proposition above is surprisingly simple:

Suppose $a_{n+1}\neq a_n$ infinitely often (otherwise the result is trivial). Notice that $$\Big(\frac{a_{n+1}}{a_n}\Big)^n=\left(\Big(1+\frac{a_{n+1}-a_n}{a_n}\Big)^{\tfrac{a_n}{a_{n+1}-a_n}}\right)^{\tfrac{n(a_{n+1}-a_n)}{a_n}} $$ Taking logarithms yields $$\frac{a_n}{n}\log\Big(\big(\frac{a_{n+1}}{a_n}\big)^n\Big)=(a_{n+1}-a_n)\log\left(\Big(1+\frac{a_{n+1}-a_n}{a_n}\Big)^{\tfrac{a_n}{a_{n+1}-a_n}}\right) $$

Condition (1) implies that $\frac{a_{n+1}}{a_n}=\frac{a_{n+1}}{n+1}\frac{n}{a_n}\frac{n}{n+1}\xrightarrow{n\rightarrow\infty}1$ and so, $\lim_n\frac{a_{n+1}-a_n}{a_n}=0$. In turn, this implies that $\log\left(\Big(1+\frac{a_{n+1}-a_n}{a_n}\Big)^{\tfrac{a_n}{a_{n+1}-a_n}}\right)\xrightarrow{n\rightarrow\infty}1$, which follows from the fact that $\lim_{x\rightarrow0}\frac1h\log(1+h)=\log'(1)=1$. Consequently $$A\log B=\lim_n(a_{n+1}-a_n)$$

As an application of the Proposition L, under the condition of the OP, define $a_n=\sqrt[n]{c_n}$. Notice that $\frac{a_n}{n}=\sqrt[n]{\frac{c_n}{n^n}}$. Since $$\lim_n\frac{c_{n+1}}{(n+1)^{n+1}}\frac{n^n}{c_n}=\lim_n\frac{c_{n+1}}{nc_n}\Big(1+\frac{1}{n}\Big)^{-n}\Big(\frac{n}{n+1}\Big)=\frac{a}{e}, $$ Cauchy-D'Lambert's theorem yields $$\lim_n\frac{a_n}{n}=\frac{a}{e}$$

Observe that $$\Big(\frac{a_{n+1}}{a_n}\Big)^n=\frac{c_{n+1}}{nc_n}\frac{n+1}{\sqrt[n+1]{c_{n+1}}}\frac{n}{n+1}\xrightarrow{n\rightarrow\infty}a\frac{e}{a}=e $$ Hence $$a_{n+1}-a_n=\sqrt[n+1]{c_{n+1}}-\sqrt[n]{c_n}\xrightarrow{n\rightarrow\infty}\frac{a}{e}\log e=\frac{a}{e} $$

  • 1
    $\begingroup$ Ok, this was way harder to understand than I thought it would be😅. Now I just want to confirm that the Cauchy-D'Lambert theorem you stated, is it the same as the modified form of the multiplicative identity of the Stolz - Cesaro theorem (this) $\endgroup$
    – Ham Lemon
    Nov 21 at 5:49
  • $\begingroup$ @HamLemon: You can prove a version of the Cauchy-D'Lambert theorem using Cesaro-Stolz theorem. But there are more direct ways to do it; also, it is a result use seen in many Calculus courses in the study of absolute convergence of series (root test and ratio test): for positive sequence $a_n$, $\liminf_n\frac{a_{n+1}}{a_n}\leq\liminf_n\sqrt[n]{a_n}\leq\limsup_n\sqrt[n]{a_n}\leq\limsup_n\frac{a_{n+1}}{a_n}$ $\endgroup$
    – Mittens
    Nov 21 at 11:53
  • $\begingroup$ @HamLemon: One thing you should notice is that the pro I presented does not used anything too advanced. Finally, if you want to make a further connection with Cesar-Stolz, you can deduce what the value of $B$ must be. $\endgroup$
    – Mittens
    Nov 21 at 16:18
  • $\begingroup$ Thank you for your explanation, I understand now🙏🏼 $\endgroup$
    – Ham Lemon
    Nov 22 at 6:20

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