How do I evaluate the limit of the following sequence $$a_n = \left(\left( 1 + \frac1n \right) \left( 1 + \frac2n \right)\cdots\left( 1 + \frac nn \right) \right)^{1/n}$$

I tried to take log and then using cauchy $1^{st}$ theorem of limits but I coudn't do it.

Could someone kindly help?


2 Answers 2


Consider that $\ln a_n$ is a sequence of Riemann sums for the function $f(x) = \ln(1 + x)$ over the interval $[0,1]$. Thus $$\lim_{n \to \infty} \ln a_n = \int_0^1 \ln(1 + x)\, dx = (1 + x)\ln(1 + x) - (1 + x)|_{x = 0}^1 = 2\ln(2) - 1$$ Now you can find $\lim_{n \to \infty} a_n$ by exponentiation.

  • $\begingroup$ Can it be done using cauchy theorem of sequences .thanks $\endgroup$
    – godonichia
    Commented Dec 16, 2014 at 20:19
  • $\begingroup$ It can be done with Cauchy's second theorem, but I don't think it can be done with the first theorem. $\endgroup$
    – kobe
    Commented Dec 16, 2014 at 20:35

Rewrite $\displaystyle a_n = \prod\limits_{k=1}^{n} \left(1+\frac{k}{n}\right)^{1/n} $

Since, $\displaystyle \lim\limits_{n\to \infty} b_n^{1/n} = \lim\limits_{n\to \infty} \frac{b_{n+1}}{b_n}$ (that is if the limit exists)

We may write:

$\displaystyle \begin{align} \lim\limits_{n\to \infty} a_n =\lim\limits_{n\to \infty} \frac{\prod\limits_{k=1}^{n+1} \left(1+\frac{k}{n+1}\right)}{\prod\limits_{k=1}^{n} \left(1+\frac{k}{n}\right)} &= \lim\limits_{n \to \infty} \frac{\prod\limits_{k=1}^{n+1} (n+k+1)}{\prod\limits_{k=1}^{n} (n+k)}.\frac{(n+1)^{-(n+1)}}{n^{-n}} \tag{1}\\ &=\lim\limits_{n\to \infty} \frac{(n+2)\cdots(2n+2)}{(n+1)\cdots(2n)}.\frac{\left(1+\frac{1}{n}\right)^{-n}}{n+1} \tag{2}\\&= \lim\limits_{n\to \infty} 2(2n+1).\frac{\left(1+\frac{1}{n}\right)^{-n}}{n+1} \tag{3} \\&= \frac{4}{e}\end{align}$

2nd Approach:

Write $\displaystyle a_n = \left(\binom{2n}{n}\frac{n!}{n^n}\right)^{1/n}$

Using the identity $\displaystyle \sum\limits_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n}$ we make an application of Cauchy-Schwarz Ineqality

$$\displaystyle (n+1)\binom{2n}{n} = \left(\sum\limits_{k=0}^{n} 1^2\right)\left(\sum\limits_{k=0}^{n} \binom{n}{k}^2\right) \ge \left(\sum\limits_{k=0}^{n} \binom{n}{k}\right)^2 = 4^n$$

Thus, $\displaystyle \frac{4^n}{n+1} \le \binom{2n}{n} \le 4^n \implies \lim\limits_{n \to \infty} \frac{4}{\sqrt[n]{n+1}} = \lim\limits_{n \to \infty} \binom{2n}{n}^{1/n} = 4$ (form, squeeze theorem).

Since, $\displaystyle \lim\limits_{n \to \infty} \frac{(n!)^{1/n}}{n} = e^{-1}$ we get $\displaystyle \lim\limits_{n \to \infty} a_n = \frac{4}{e}$

  • $\begingroup$ Can you please elaborate bit more about last step $\endgroup$
    – godonichia
    Commented Dec 16, 2014 at 20:39
  • $\begingroup$ @godonichia in the last step I used $\lim\limits_{n\to \infty} \left(1+\frac{1}{n}\right)^{n} = e$. $\endgroup$
    – r9m
    Commented Dec 16, 2014 at 20:40
  • $\begingroup$ No from 4th last to 3rd last $\endgroup$
    – godonichia
    Commented Dec 16, 2014 at 20:49
  • $\begingroup$ @godonichia I edited ! if its not clear yet please indicate which line I need to elaborate on ! :) $\endgroup$
    – r9m
    Commented Dec 16, 2014 at 20:58
  • $\begingroup$ In 1st step from where did the term after product came from $\endgroup$
    – godonichia
    Commented Dec 16, 2014 at 21:03

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