In the article "Interesting Limit Problem: Arbitrary Scaling/Bounding", the reader "Zuijin Daole" proposed a novel solution. However, that reader did not write it particularly clearly, and more importantly, some of the techniques involved seemed to be ones I had never seen before. Consequently, I analyzed it myself and have provided the following explanation.
A Nonsensical Result
Suppose we want to calculate the sum of the 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 in a sequence, such as A_k, \, k=0,1,2,\dots in the expression above. However, some people dislike this and prefer to use superscripts to represent the terms of the sequence, writing the series as: \sum_{k=0}^n \binom{n}{k}\frac{A^k}{n^k} Readers might object: "Isn’t this just messing around? Won’t it be confused with the k-th power of n in the denominator?" But that person might go even further and write the series as: \sum_{k=0}^n \binom{n}{k}\frac{A^k}{n^k}=\left(1+\frac{A}{n}\right)^n Do you see? He simply treats A as if it were a number! This is too much nonsense—what exactly is A? One might expect such a student to be kicked out of the classroom by their teacher.
However, looking at it from another perspective, it might not be entirely impermissible.
We can treat A as a linear operator. After it acts on a real number x for k times, it produces A_k x; that is, A^k x = A_k x. Then, if we apply \left(1+\frac{A}{n}\right)^n to the real number 1, we naturally obtain the original series. What is the benefit of doing this? If we are seeking 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 result is based on A being a linear operator. Furthermore, since only one operator is introduced, there are no issues with non-commutative algebra; therefore, all operational laws are preserved!
A Victory of Nonsense
Now let us see what applications this nonsensical 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 seems to be no simple trick to solve this directly.
However, using our nonsensical result, it becomes quite 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 happens to be the correct final result!
Readers can construct other examples as well; I have only thought of this simple one.
Now, let’s look back at the solution by the reader "Zuijin Daole" in the article "Interesting Limit Problem: Arbitrary Scaling/Bounding". The original problem was: \lim_{n\to\infty} \frac{1^n + 2^n +\dots + n^n}{n^n} For the summation of the series in the numerator, we have (please refer to the Wikipedia entry for 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 becomes: \sum_{k=0}^{n} k^n = (B+n+1)^n Therefore: \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! Thus: \lim_{n\to\infty} \frac{1^n + 2^n +\dots + n^n}{n^n} = \frac{e}{e-1}
Subtleties
I have not yet fully grasped the subtlety of the one-to-one correspondence between sequences and powers of operators. I have a vague feeling that it must be closely related to the method of generating functions. If I obtain further results, I will share them with everyone.
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