This question has been asked before but I did not understand the accepted answer and it was 8 years old so I did not want to comment on it.

I want to understand what is incorrect in my working of the following- $$ \lim \limits_{n \to \infty} (n!)^{\frac{1}{n}} = \lim\limits_{n \to \infty}[n^{\frac{1}{n}}\cdot (n-1)^{\frac{1}{n}} \cdot (n-2)^{\frac{1}{n}} \cdot \ldots\cdot 1^{\frac{1}{n}}] \tag{1}$$

Now, the product rule of limits says, $$ \lim\limits_{x \to a}[f(x) \cdot g(x)] = \lim\limits_{x \to a} f(x) \cdot \lim\limits_{x \to a} g(x)$$

Let $f(n) = n^{\frac{1}{n}}$, and the remaining be $g(n)$. Using this in $(1)$, I get-

$$\lim\limits_{n \to \infty}(n!)^{\frac{1}{n}} = \lim\limits_{n \to \infty}n^{\frac{1}{n}} \cdot \lim\limits_{n \to \infty}[(n-1)^{\frac{1}{n}} \cdot (n-2)^{\frac{1}{n}} \cdot \ldots \cdot 1^{\frac{1}{n}}]$$

Repeating the step and continuously breaking the rightmost limit using product law of limits, I get-

$$\lim\limits_{n \to \infty}(n!)^{\frac{1}{n}} = \lim\limits_{n \to \infty}n^{\frac{1}{n}} \cdot \lim\limits_{n \to \infty}(n-1)^{\frac{1}{n}} \cdot \lim\limits_{n \to \infty}(n-2)^{\frac{1}{n}} \cdot \ldots \cdot \lim\limits_{n \to \infty}1^{\frac{1}{n}}$$

Now, I know the limit of each of these individual terms is $1$. So, I write it as- $$\lim\limits_{n \to \infty}(n!)^{\frac{1}{n}} = 1 \cdot 1 \cdot 1 \cdot \ldots \cdot 1 = 1$$

I know this answer is incorrect as the correct limit is $+\infty$, which can be seen through Stirling's Approximation, but I don't know what is incorrect in my working.

  • 1
    $\begingroup$ The number of breaking the limit is $n$, which depends on $n$, which approaches to $\infty$. This cannot be done. $\endgroup$
    – Riemann
    Oct 30, 2022 at 13:18
  • 13
    $\begingroup$ The product rule for limits holds only for a product of a (finite and) fixed number of factors. $\endgroup$ Oct 30, 2022 at 13:18
  • $\begingroup$ You cannot treat the limit of an infinite product as the products of the limits. By the same reasoning, you would have the limit of the Riemann sum of a function always result in zero, which is false. You have to treat the function as a whole. $\endgroup$
    – pnguib
    Oct 30, 2022 at 13:20
  • 2
    $\begingroup$ You could apply the same line of reasoning to $\displaystyle (1+1/n)^n\color{red}{\to} \underbrace{1\cdot \ldots\cdot 1}_{n}=1$, which should really be $\ldots \to e$. As the above comments have noted, you cannot generally apply $\lim (a\cdot b)=\lim a\cdot \lim b$ infinitely many times in one go. $\endgroup$
    – Jam
    Oct 30, 2022 at 13:27
  • $\begingroup$ It is a worthwhile exercise, in thinking through the various ways this doesn't work, to address the question: what is the generic term in the ". . ." in the expression $$ \lim\limits_{n \to \infty}n^{\frac{1}{n}} \cdot \lim\limits_{n \to \infty}(n-1)^{\frac{1}{n}} \cdot \lim\limits_{n \to \infty}(n-2)^{\frac{1}{n}} \cdot \ldots \cdot \lim\limits_{n \to \infty}1^{\frac{1}{n}}$$ supposed to be? (In all the expressions before that, it is easy to explain. In that one, no so much, since $n$ does not exist outside the limit.) $\endgroup$
    – JBL
    Oct 30, 2022 at 13:55

1 Answer 1


You cannot use the product rule as the number of terms in the sequence is not fixed. Instead, if you consider $a_n$ as the sequence, then as the terms of the sequence are positive you have: $a_n = \text{exp}(\log(a_n)) = \exp(\frac{1}{n}\sum^{n}_{i=2}\log(i))$. Now since $\log$ is strictly increasing we have(when $n$ is even),

$a_n =\exp(\frac{1}{n}\sum^{n}_{i=2}\log(i)) \geq \exp(\frac{1}{n}\sum^{n/2}_{i=2}\log(i)) + \frac{1}{2}\log(\frac{n}{2})) \geq \exp(\frac{1}{n}\sum^{n/2}_{i=2}\log(2)) + \frac{1}{2}\log(\frac{n}{2}))\geq \sqrt{\frac{n}{2}}\exp(1/6), n \geq 4$.

Hence the sequence is unbounded.


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