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Let $(a_n)_{n\geq 1}$ and $(b_n)_{n\geq 1}$ be positive real sequences such that $$\lim_{n\to\infty}\frac{a_{n+1}-a_n}n=a\in \mathbb R_{>0}\qquad\text{and}\qquad \lim_{n\to\infty}\frac{b_{n+1}}{nb_n}=b\in \mathbb R_{>0}.$$ Compute $$\lim_{n\to\infty}\left(\frac{a_{n+1}}{\sqrt[n+1]{b_{n+1}}}-\frac{a_n}{\sqrt[n]{b_n}}\right).$$

This is W5 of József Wildt International Mathematical Competition, 2020. I have made some progresses and got stuck.

Using Cauchy's criterion, I have obtained that $\sqrt[n]{\frac{b_{n+1}}{b_1n!}}\to b$ and thus $$\sqrt[n]{\frac{b_n}{b_1n!}}=\sqrt[n]{\frac{b_{n+1}}{b_1n!}}\cdot\sqrt[n]{\frac{nb_n}{b_{n+1}}}\cdot\sqrt[n]{\frac1n}\to b,\qquad\text{as}\ \ n\to\infty.$$ Stirling's formula implies that $\lim_{n\to\infty}\frac{\sqrt[n]{b_n}}n=\frac be$. Now it seems resonable to guess that $$\lim_{n\to\infty}\left(\frac{a_{n+1}}{\sqrt[n+1]{b_{n+1}}}-\frac{a_n}{\sqrt[n]{b_n}}\right)=\lim_{n\to\infty}\left(\frac{\frac{a_{n+1}}{n}}{\frac{\sqrt[n+1]{b_{n+1}}}{n}}-\frac{\frac{a_n}n}{\frac{\sqrt[n]{b_n}}n}\right)\overset{?}{=}\lim_{n\to\infty}\frac eb\frac{a_{n+1}-a_n}n=\frac{ae}b\tag{1};$$ or another possibility $$\lim_{n\to\infty}\left(\frac{a_{n+1}}{\sqrt[n+1]{b_{n+1}}}-\frac{a_n}{\sqrt[n]{b_n}}\right)=\lim_{n\to\infty}\left(\frac{\frac{a_{n+1}}{n+1}}{\frac{\sqrt[n+1]{b_{n+1}}}{n+1}}-\frac{\frac{a_n}n}{\frac{\sqrt[n]{b_n}}n}\right)\overset{?}{=}\frac eb\lim_{n\to\infty}\left(\frac{a_{n+1}}{n+1}-\frac{a_n}n\right),\tag{2}$$ where the limit $\lim_{n\to\infty}\left(\frac{a_{n+1}}{n+1}-\frac{a_n}n\right)$ can be obtained by $$\frac{a_{n+1}}{n+1}-\frac{a_n}n=\frac{a_{n+1}-a_n}n+\frac{a_{n+1}}{n+1}-\frac{a_{n+1}}n=\frac{a_{n+1}-a_n}n-\frac{a_{n+1}}{n(n+1)}$$ and Cesaro-Stolz: $\lim_{n\to\infty}\frac{a_{n+1}}{n(n+1)}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{n(n+1)-(n-1)n}=\frac a2$, so $\lim_{n\to\infty}\left(\frac{a_{n+1}}{n+1}-\frac{a_n}n\right)=a-\frac a2=\frac a2$. In $(1)$ and $(2)$ above, I don't know the eact reasons for both '$?$', I just guess them.

Therefore, when I looked at the same limit from two different points of view, namely $(1)$ and $(2)$, I got different results, which makes me confused. I wonder which one is correct and what is the rigorous proof.

Could someone help me to finish this problem?

Feng
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2 Answers2

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Let $c_n=a_n/\sqrt[n] {b_n} $ and then we have $$\frac{c_{n+1}}{c_n}=\frac{a_{n+1}}{a_n}\cdot\frac{\sqrt[n]{b_n}/n}{\sqrt[n+1]{b_{n+1}}/(n+1)}\cdot\frac{n}{n+1}$$ so that the ratio $c_{n+1}/c_n$ tends to same limit as that of $a_{n+1}/a_n$.

Using Cesaro-Stolz we see that $2a_n/n^2\to a$. And hence $a_{n+1}/a_n\to 1$ and therefore $c_{n+1}/c_n\to 1$.

We can next observe that $$c_{n+1}-c_n=c_n\cdot\left(\frac{c_{n+1}}{c_n}-1\right)=\frac{a_n/n} {\sqrt[n] {b_n} /n} \cdot\frac{c_{n+1}/c_n-1}{\log(c_{n+1}/c_n)}\cdot\log\frac{c_{n+1}}{c_n}$$ And then desired limit is same as that of $$\frac{e} {b}\cdot \frac{a_n} {n^2}\cdot n\log\frac{c_{n+1}}{c_n}$$ The limit of above expression is same as that of $$\frac{ae} {2b} \left(n\log\frac{a_{n+1}}{a_n}+\log\frac{b_n\sqrt[n+1]{b_{n+1}}}{b_{n+1}}\right) $$ The second term in parentheses can be written as $$\log\left(\frac{nb_n} {b_{n+1}}\cdot\frac{\sqrt[n+1]{b_{n+1}}}{n+1}\cdot\frac{n+1}{n}\right) $$ and this tends to $$\log((1/b)(b/e)(1))=-1$$ and the first term can be written as $$n\cdot\frac{\log (a_{n+1}/a_n)} {(a_{n+1}/a_n)-1}\cdot\left(\frac{a_{n+1}}{a_n}-1\right)$$ whose limit is same as that of $$\frac{n^2}{a_n}\cdot\frac{a_{n+1}-a_n}{n}$$ which is $(2/a) \cdot a=2$. The desired limit in question is thus $\dfrac{ae} {2b} $.

Paramanand Singh
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2

Here is a rather nice result which helps in this kinds of problems:

Theorem (O. Carja): Suppose $(a_n:n\in\mathbb{N})$ is a sequence of positive numbers such that
(i) $\lim_n\frac{p_n}{n}=p>0$
(ii) $\lim_n\big(\frac{p_{n+1}}{p_n}\big)^n=q\in\overline{\mathbb{R}}_+$.
Then $$L=\lim_n(p_{n+1}-p_n)$$ exists and $L=p\log q$.

A simple proof using elementary Calculus is discussed here.

To apply this result to the problem of the OP define $c_n=\frac{a_n}{\sqrt[n]{b_n}}$

The first condition in the problem of the OP implies that $$\frac{a_{n+1}-a_n}{(n+1)^2-n^2}=\frac{a_{n+1}-a_n}{n}\frac{n}{2n+1}\xrightarrow{n\rightarrow\infty}\frac{a}{2}\in(0,\infty)$$ In turn, this gives $$\frac{a_n}{n^2}\xrightarrow{n\rightarrow\infty}\frac{a}{2}$$ by Cèsaro-Stolz. Thus $$\frac{a_{n+1}}{a_n}=\frac{a_{n+1}}{(n+1)^2}\frac{n^2}{a_n}\frac{(n+1)^2}{n^2}\xrightarrow{n\rightarrow\infty}1$$

The condition $\frac{b_{n+1}}{nb_n}\xrightarrow{n\rightarrow\infty}b$ implies that $$\frac{b_{n+1}}{(n+1)^{n+1}}\frac{n^n}{b_n}=\frac{b_{n+1}}{nb_n}\frac{1}{\Big(1+\frac{1}{n}\Big)}\frac{1}{\Big(1+\frac1n\Big)^n}\xrightarrow{n\rightarrow\infty}\frac{b}{e}$$ Hence $$\frac{\sqrt[n]{b_n}}{n}\xrightarrow{n\rightarrow\infty}\frac{b}{e}$$

Consequently \begin{align} \frac{c_n}{n}=\frac{a_n}{n\sqrt[n]{b_n}}=\frac{a_n}{n^2}\frac{n}{\sqrt[n]{b_n}}\xrightarrow{n\rightarrow\infty}\frac{a}{2}\frac{e}{b}\tag{1}\label{one} \end{align}

Now $$\log\big(\big(\frac{a_{n+1}}{a_n}\big)^n\Big)=\frac{\log\Big(1+\big(\frac{a_{n+1}}{a_n}-1\big)\Big)}{\frac{a_{n+1}}{a_n}-1}\big(\frac{a_{n+1}-a_n}{n}\big)\frac{n^2}{a_n}\xrightarrow{n\rightarrow\infty}\log'(1)a\frac2a=2$$

Hence \begin{align} \Big(\frac{c_{n+1}}{c_n}\Big)^n=\Big(\frac{a_{n+1}}{a_n}\Big)^n\frac{n b_n}{b_{n+1}}\frac{\sqrt[n+1]{b_{n+1}}}{n+1}\frac{n+1}{n}\xrightarrow{n\rightarrow\infty}e^2\frac{1}{b}\frac{b}{e}=e\tag{2}\label{two} \end{align}

(1) and (2) implies that $$\lim_n(c_{n+1}-c_n)= \frac{a}{2}\frac{e}{b}\log e=\frac{ae}{2b}$$

Mittens
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