Problem Statement:
Let $\displaystyle a_{n} = \frac{\left(2n\right)!}{\left(n!\right)^{2}4^{n}}$.
$$
\mbox{Compute the limit:}\quad
\lim_{n \to \infty}\left\{n\left[\frac{a_{n + 1}}{a_{n}} – \left(\frac{n}{n + 1}\right)^{1/3} \right]\right\}
$$
Hence, show that $a_{n} \to 0$.
Context:
- No using L'Hospital's Rule and the concept of functions or continuity as that would be circular reasoning.
- Just the concept(s) of formal definition of limits, arithmetic laws of limits, squeeze theorem, and general comparison of limits, sequence ratio/root tests.
Sequence Ratio test:
Suppose $\{a_n\}$ is a sequence of real numbers such that
\begin{align*}
\lim_{n\to \infty} \frac{|a_{n+1}|}{|a_n|} = L
\end{align*}
Where $0\leq L < 1$, then $\lim_{n \to \infty}a_n =0$
Simplified form of the expression:
\begin{align*}
\lim_{n \to \infty} n\left[ \frac{n+1/2}{n+1} – \left(\frac{n}{n+1}\right)^{\frac{1}{3}}\right]
\end{align*}
Have tried to to use the sequence ratio test by
\begin{align*}
b_n := n\left[ \frac{n+1/2}{n+1} – \left(\frac{n}{n+1}\right)^{\frac{1}{3}}\right]
\end{align*}
But, this gave inconclusive results of $\lim_{n\to \infty }\frac{b_{n+1}}{b_n} = \frac{0}{0}$.
Manipulation of the limit expression using the arithmetic laws of limits is blocked by the $n$ term as $\lim_{n\to \infty} n$ does not exist.
How does one go about computing the limit of sequences of this variety?
Best Answer
Note that: $$n\left[\frac{n+1/2}{n+1}-\left(\frac{n}{n+1}\right)^{\frac{1}{3}}\right] =\frac{n}{n+1}\left[(n+1/2)-\sqrt[3]{n(n+1)^2}\right].$$ $$\lim_{n\to\infty}n\left[\frac{n+1/2}{n+1}-\left(\frac{n}{n+1}\right)^{\frac{1}{3}}\right] =\lim_{n\to\infty}\left[(n+1/2)-\sqrt[3]{n(n+1)^2}\right]\\ =\lim_{n\to\infty}\frac{(n+1/2)^3-n(n+1)^2}{(n+1/2)^2+(n+1/2)\sqrt[3]{n(n+1)^2}+[n(n+1)^2]^{\frac23}}\\ =\lim_{n\to\infty}\frac{-\frac12n^2-\frac14n+\frac18}{(n+1/2)^2+(n+1/2)\sqrt[3]{n(n+1)^2}+[n(n+1)^2]^{\frac23}}\\ =\lim_{n\to\infty}\frac{-\frac12-\frac1{4n}+\frac1{8n^2}}{(1+1/(2n))^2+(1+1/(2n))\sqrt[3]{(1+1/n)^2}+[(1+1/n)^2]^{\frac23}}\\ =-\frac16.$$
Added： $$\left[(n+1/2)-\sqrt[3]{n(n+1)^2}\right] =n\left[1+1/(2n)-\left(1+\frac1n\right)^{\frac23}\right]\\ =n\left[1+1/(2n)-\left(1+\frac23\cdot\frac1n+o(1/n)\right)\right]\\ =-\frac16+o(1)\to-\frac16\quad n\to\infty.$$