This post comes mostly from the youtube video by BlackPenRedPen found here:

This in turn comes from – details and links can be found in the original video

In this post we will have a look at a complicated-looking limit that has an interesting solution. Here it is:

\lim_{n \rightarrow \infty} ( \frac{n!}{n^n})^{\frac{1}{n}}

This looks pretty daunting – but we will break the solution down into sections:

  • taking the logarithms and rearranging
  • recognising something familiar
  • finding the numerical value


Step 1: Taking the Logarithm

The first step here is to take the logarithm, a generally useful trick when applying limits. First we assign the variable L to the limit (so that we can solve for it in the end). Now lets do some algebra:

L = \lim_{n \rightarrow \infty} ( \frac{n!}{n^n})^{\frac{1}{n}}

\ln(L) = \ln(\lim_{n \rightarrow \infty} ( \frac{n!}{n^n})^{\frac{1}{n}})

Noting that the natural logarithm \ln is a continuous function and therefore we can take the limit outside of the function:

\ln(L) =  \lim_{n \rightarrow \infty} \ln( (\frac{n!}{n^n})^{\frac{1}{n}})

Next we can use the logarithm laws to bring down the exponent:

\ln(L) =  \lim_{n \rightarrow \infty}  \frac{1}{n} \ln(\frac{n!}{n^n})

Alright, now we have taken the logarithm, step 1 is complete.

Step 2: Rearranging the expression and recognising the form of the integral

Now we turn our attention to the expression \frac{n!}{n^n}. Breaking this down a bit more we can get closer to solving this. We know that n! = n*(n-1)*(n-2) \dots 1 and that n^n = n*n*n \dots n.

So we can rewrite:

\ln(\frac{n!}{n^n}) = \ln(\frac{n}{n} . \frac{n-1}{n} . \frac{n-2}{n}\dots \frac{1}{n})

Then using log laws and changing the order we have:

\ln(\frac{n!}{n^n}) = \ln(\frac{1}{n}) +\ln(\frac{2}{n}) + \ln(\frac{3}{n}) +\dots + \ln(\frac{n-2}{n}) + \ln(\frac{n-1}{n}) +\ln(\frac{n}{n})

Next we put this back into our full expression to get:

\ln(L) =  \lim_{n \rightarrow \infty}  \frac{1}{n} (\ln(\frac{1}{n}) +\ln(\frac{2}{n}) + \ln(\frac{3}{n}) +\dots + \ln(\frac{n-2}{n}) + \ln(\frac{n-1}{n}) +\ln(\frac{n}{n}))

\ln(L) =  \lim_{n \rightarrow \infty}  \frac{1}{n}\sum_i^n \ln(\frac{1}{n})

But this looks a lot like the definition of the Riemann integral:

\lim_{n \rightarrow \infty} \frac{1}{n} \sum_i^n f(x_i)

where we have split the interval [0,1] into increments, x_i = \frac{i}{n} and f(x) = \ln(x).

So we can therefore rewrite our expression above as:

\ln(L) =  \int_0^1 \ln(x) dx

Now all that we need to do is find the value of the integral

Step 3: Evaluating the integral and solving for L

The last thing we need to do is solve this integral – we will do this by parts.

The 2 functions f,g and their derivatives (f',g') are as follows:

f = ln(x), g = x, f' = \frac{1}{x}, g' = 1

which gives us:

x\ln(x)|^1_0 - \int_0^1 x \frac{1}{x} dx

Looking at the first term we have:

1.\ln(1) - \lim_{x \rightarrow 0^+} x.ln(x)

= 0 - \lim_{x \rightarrow 0^+} x.ln(x)

Now to handle this limit, which is in the indeterminate form – as usual we will rewrite it and then use L’Hospital’s rule to evaluate it:

\lim_{x \rightarrow 0^+} x.ln(x) = 0 \times -\infty

\lim_{x \rightarrow 0^+} x.ln(x) = \lim_{x \rightarrow 0^+} \frac{ln(x)}{\frac{1}{x}} = \frac{-\infty}{\infty}

Using L’Hospital’s rule and differentiating the numerator and the denominator and taking their respective limits we get:

\lim_{x \rightarrow 0^+} x.ln(x) = \lim_{x \rightarrow 0^+} \frac{ln(x)}{\frac{1}{x}} = \lim_{x \rightarrow 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \rightarrow 0^+} -x = 0 

This gives us the first term as 0.

Clearly the second term, -\int_0^1 x \frac{1}{x} dx =-\int_0^1 1 dx  = -x|^1_0 = -(1-0),

is   -1.

Therefore our integral evaluates to -1.

Now all that remains is to rearrange our equation and solve for L

\ln(L) =  \int_0^1 \ln(x) dx = -1

\ln(L) = -1

L = e^{-1}

and that’s it, we are done:

\lim_{n \rightarrow \infty} ( \frac{n!}{n^n})^{\frac{1}{n}} = e^{-1}



How clear is this post?