By 苏剑林 | May 26, 2015
In the article "An Interesting Limit Problem: Scaling at Will", the reader "Recently Fell Down" proposed a novel solution. However, this reader did not write it very clearly. More importantly, some of the techniques involved seem to be things I haven't encountered before. So, I analyzed it myself and provided the following explanation.
Suppose we want to find the sum of a series:
\[ \sum_{k=0}^n \binom{n}{k}\frac{A_k}{n^k} \]where $A_0=1$. Generally, we use subscripts to denote different terms, such as $A_k, k=0,1,2,\dots$ in the above formula. However, some people don't like this; they prefer using exponents to represent the terms of the sequence. They write the series above as:
\[ \sum_{k=0}^n \binom{n}{k}\frac{A^k}{n^k} \]Readers might object: "Isn't this just fooling around? Wouldn't this be confused with the $k$-th power of $n$ in the denominator?" But that person goes even further and writes the series as:
\[ \sum_{k=0}^n \binom{n}{k}\frac{A^k}{n^k}=\left(1+\frac{A}{n}\right)^n \]See that? He simply treated $A$ as a number! It's complete nonsense. What is $A$? A kid like that would probably be kicked out of the classroom by his teacher.
But looking at it from another perspective, it might just work.
Let us regard $A$ as a linear operator that, when applied $k$ times to a real number $x$, produces $A_k x$. That is, $A^k x = A_k x$. Then, applying $\left(1+\frac{A}{n}\right)^n$ to the real number 1 naturally yields the original series. What is the benefit of doing this? If we are finding the limit as $n \to \infty$, some interesting things happen:
\begin{aligned}&\lim_{n\to\infty} \sum_{k=0}^n \binom{n}{k}\frac{A_k}{n^k}\\ =&\lim_{n\to\infty} \left(1+\frac{A}{n}\right)^n\\ =&e^A\\ =&\sum_{k=0}^{\infty} \frac{1}{k!}A^k\\ =&\sum_{k=0}^{\infty} \frac{1}{k!}A_k\end{aligned}Here we have achieved a transformation from one series to another! This is based on the result that $A$ is a linear operator. Furthermore, since only one operator is introduced, there are no issues with non-commutative algebra; thus, all operational laws are preserved!
Now let's see what applications this "nonsense" result has. Suppose $A_k=\frac{k!}{2^k}$, and we want to calculate the limit of our original series, which is:
\[ \lim_{n\to\infty}\sum_{k=0}^n \frac{n!}{(n-k)!}\frac{1}{(2n)^k} \]There doesn't seem to be any simple trick to do this.
However, using our "nonsense" result, it becomes relatively simple:
\begin{aligned}&\lim_{n\to\infty} \sum_{k=0}^n \binom{n}{k}\frac{A_k}{n^k}\\ =&e^A\\ =&\sum_{k=0}^{\infty} \frac{1}{k!}A_k\\ =&\sum_{k=0}^{\infty} \left(\frac{1}{2}\right)^k\\ =&2\end{aligned}This is exactly the correct final result!
Readers can likewise construct other examples. Being somewhat dull, I could only think of this simple one.
Now, let's look at the solution by the reader "Recently Fell Down" in the article "An Interesting Limit Problem: Scaling at Will". The original problem was:
\[ \lim_{n\to\infty} \frac{1^n + 2^n +\dots + n^n}{n^n} \]For the summation in the numerator, we have (please refer to the Wikipedia page on "Bernoulli numbers"):
\[ \sum_{k=0}^{m-1} k^n = \frac{1}{n+1}\sum_{k=0}^n \binom{n+1}{k} B_k m^{n+1-k} \]Thus we have:
\[ \sum_{k=0}^{n} k^n = \frac{1}{n+1}\sum_{k=0}^n \binom{n+1}{k} B_k (n+1)^{n+1-k} = \sum_{k=0}^n \binom{n+1}{k} B_k (n+1)^{n-k} \]According to the "nonsense" above, treating $B$ as an operator where $B^k=B_k$, the above expression is:
\[ \sum_{k=0}^{n} k^n = (B+n+1)^n \]Thus:
\begin{aligned}&\lim_{n\to\infty} \frac{1^n + 2^n +\dots + n^n}{n^n}\\ =&\lim_{n\to\infty} \frac{(B+n+1)^n}{n^n}\\ =&\lim_{n\to\infty} \left(1+\frac{B+1}{n}\right)^n\\ =& e^{B+1} =e\times e^B\end{aligned}Note that:
\[ e^B=\sum_{k=0}^{\infty} \frac{1}{k!}B_k=\frac{1}{e-1} \]This is exactly the result defined by the generating function of Bernoulli numbers! Therefore:
\[ \lim_{n\to\infty} \frac{1^n + 2^n +\dots + n^n}{n^n}=\frac{e}{e-1} \]I haven't yet fully grasped the subtleties of mapping a sequence to the powers of an operator. I vaguely feel that it must be closely related to the generating function method. If I have further results, I will come back and share them with everyone.