Limit of a sequence by Cauchy second test

Question

The sequence is

$$ \left[ \bigg(1+\frac{1}{n}\bigg)\bigg(1+\frac{2}{n}\bigg)\bigg(1+\frac{3}{n}\bigg)\cdots\bigg(1+\frac{n}{n}\bigg) \right]^{1/n} $$

as $n$ goes to infinity. By Cauchy second test it's pretty clear that it's limit will be equal to $1+\frac{n}{n}$ which is $2$. But the answer is $4/e$. I don't know what part I'm doing wrong.

The Cauchy test only says if it converges or not. Limit calculation is a different thing. — Sarvesh Ravichandran Iyer, Jan 26 '17 at 04:42
Perhaps he means The sequence is $\displaystyle \left[ \left(1+\frac{1}{n}\right)\cdot \left(1+\frac{2}{n}\right)\cdot \left(1+\frac{3}{n}\right)\cdot \cdots\cdot \left(1+\frac{n}{n}\right) \right]^{1/n} $ — juantheron, Jan 26 '17 at 05:34
Taking log, you will see a Riemann sum of some nice function. — Sangchul Lee, Jan 26 '17 at 05:48

score 3 · Answer 1 · answered Jan 26 '17 at 05:57

We have $a_n=\left[ \left(1+\frac{1}{n}\right).\left(1+\frac{2}{n}\right).\left(1+\frac{3}{n}\right)+\cdots.\left(1+\frac{n}{n}\right) \right]^{1/n}$ then $\log a_n=\frac1n\displaystyle\sum_{k=1}^n\log(1+\frac{k}{n})$.

Now $\displaystyle\lim_{n\to\infty}\log a_n=\int_0^1\log(1+x)dx=\log4-1$,then $\displaystyle\lim_{n\to\infty}a_n=e^{\log4-1}=\frac4e$

score 2 · Answer 2 · answered Jan 26 '17 at 05:57

With

$$a_n = \left[ \bigg(1+\frac{1}{n}\bigg)\bigg(1+\frac{2}{n}\bigg)\bigg(1+\frac{3}{n}\bigg)\cdots\bigg(1+\frac{n}{n}\bigg) \right]^{1/n},$$

Cauchy's second limit test yields

$$\lim_{n \to \infty} a_n^{1/n} = \lim_{n \to \infty}\frac{a_{n+1}}{a_n} \\ =\lim_{n \to \infty} \left(\frac{n}{n+1} \right)^n\prod_{k=1}^{n}\frac{n+1+k}{n+k}\left(1 + \frac{n+1}{n+1}\right) \\ = \lim_{n \to \infty} \frac{1}{(1 + 1/n)^n}\frac{2n+1}{n+1}\left(1 + \frac{n+1}{n+1}\right) \\ =\frac{4}{e}$$

score 1 · Answer 3 · edited Apr 13 '17 at 12:21

we can write $$\left[ \bigg(1+\frac{1}{n}\bigg)\bigg(1+\frac{2}{n}\bigg)\bigg(1+\frac{3}{n}\bigg)\cdots\bigg(1+\frac{n}{n}\bigg) \right] = n!\binom{2n}{n}\cdot \frac{1}{n^n}$$

Using $$\binom{2n}{n}\leq \binom{2n}{0}+\binom{2n}{1}+\cdots \cdots \binom{2n}{2n}=2^{2n} = 4^n$$

And Using Power mean Inequality $$\frac{\binom{n}{0}^2+\binom{n}{0}^2+\cdot \cdots \binom{n}{n}^2}{n+1}\geq \bigg[\frac{\binom{n}{0}+\binom{n}{1}+\cdots \cdots \binom{n}{n}}{n+1}\bigg]^2 = \frac{4^n}{(n+1)}$$

So $$\binom{2n}{n}=\binom{n}{0}^2+\binom{n}{0}^2+\cdot \cdots \binom{n}{n}^2>\frac{4^n}{n+1}$$

So $$\frac{4^n}{n+1}<\binom{2n}{n}\leq 4^n\Rightarrow \left(\frac{4}{(n+1)^{\frac{1}{n}}}\right)<\binom{2n}{n}^{\frac{1}{n}}\leq 4$$

So using Sequeeze Theorem $$\lim_{n\rightarrow \infty}\left(\frac{4}{(n+1)^{\frac{1}{n}}}\right)<\lim_{n\rightarrow \infty}\binom{2n}{n}^{\frac{1}{n}}\leq \lim_{n\rightarrow \infty}4$$

So we get $\displaystyle \lim_{n\rightarrow \infty}\binom{2n}{n}^{\frac{1}{n}} = 4$ and Since, $\displaystyle \lim\limits_{n \to \infty} \frac{(n!)^{1/n}}{n} = e^{-1}$ we get $$\displaystyle \lim\limits_{n \to \infty}\left[ \bigg(1+\frac{1}{n}\bigg)\bigg(1+\frac{2}{n}\bigg)\bigg(1+\frac{3}{n}\bigg)\cdots\bigg(1+\frac{n}{n}\bigg) \right]^{\frac{1}{n}} = \frac{4}{e}$$

Limit of a sequence by Cauchy second test

3 Answers3