The Victory of Nonsense: Introducing Operators into Series Summation

胡闹的胜利：将算子引入级数求和 苏剑林 · 2015-05-26 · Mathematics limit series operator

In the article "An Interesting Problem on Limits: Bounding as You Please", a reader named "最近倒了" ("recently fell over") proposed a novel solution. However, this reader's writing was not particularly clear, and more importantly, some of the tricks used seem to be things the author hadn't seen before. So I worked through it myself and offer the following explanation.

A Nonsensical Result

Suppose we want to evaluate the series sum

$$\sum_{k=0}^n \binom{n}{k}\frac{A_k}{n^k}$$

where $A_0=1$. Generally speaking, we use subscripts to label the different terms of a sequence, as with $A_k,\,k=0,1,2,\dots$ above. But some people simply prefer not to do this—they prefer to use superscripts to denote the terms of a sequence, writing the series above as

$$\sum_{k=0}^n \binom{n}{k}\frac{A^k}{n^k}$$

At this point the reader might object: isn't this just nonsense, doesn't this get confused with the k-th power of n in the denominator? But that person goes even further into nonsense, writing the series as

$$\sum_{k=0}^n \binom{n}{k}\frac{A^k}{n^k}=\left(1+\frac{A}{n}\right)^n$$

See what's happening here? He's simply treating $A$ as if it were a number! Utter nonsense—what on earth is $A$ supposed to be? A student who did this would probably get thrown out of class by the teacher.

But looking at it from another angle, it's perhaps not so unreasonable after all. more

Let's regard $A$ as a linear operator, which, when applied $k$ times to the 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 gives us back the original series. What's the benefit of doing this? If we want to compute the limit of $n\to\infty$, something rather fun happens:

$$\begin{aligned}&\lim_{n\to\infty} \sum_{k=0}^n \binom{n}{k}\frac{A_k}{n^k}\\ =&\lim_{n\to\infty} \left(1+\frac{A}{n}\right)^n\\ =&e^A\\ =&\sum_{k=0}^{\infty} \frac{1}{k!}A^k\\ =&\sum_{k=0}^{\infty} \frac{1}{k!}A_k\end{aligned}$$

Here we've achieved a transformation from one series into another! This relies on $A$ being a linear operator, and since only a single operator is introduced, there's no issue with non-commutative algebra—so all the usual rules of manipulation still hold!

The Victory of Nonsense

Now let's see what this nonsensical result is good for. Suppose $A_k=\frac{k!}{2^k}$, and we want to compute the limit of our original series, i.e.

$$\lim_{n\to\infty}\sum_{k=0}^n \frac{n!}{(n-k)!}\frac{1}{(2n)^k}$$

There doesn't seem to be any simple trick for doing this directly.

But using our nonsensical result, things become much simpler:

$$\begin{aligned}&\lim_{n\to\infty} \sum_{k=0}^n \binom{n}{k}\frac{A_k}{n^k}\\ =&e^A\\ =&\sum_{k=0}^{\infty} \frac{1}{k!}A_k\\ =&\sum_{k=0}^{\infty} \left(\frac{1}{2}\right)^k\\ =&2\end{aligned}$$

And this is exactly the correct final answer!

Readers can no doubt construct other examples of their own; being somewhat slow-witted, I could only come up with this one simple case.

Now let's go back and look at the solution given by the reader "最近倒了" in the article "An Interesting Problem on Limits: Bounding as You Please". The original problem is

$$\lim_{n\to\infty} \frac{1^n + 2^n +\dots + n^n}{n^n}$$

For the series sum in the numerator, we have (see Wikipedia's article on "Bernoulli numbers"):

$$\sum_{k=0}^{m-1} k^n = \frac{1}{n+1}\sum_{k=0}^n \binom{n+1}{k} B_k m^{n+1-k}$$

so we get

$$\sum_{k=0}^{n} k^n = \frac{1}{n+1}\sum_{k=0}^n \binom{n+1}{k} B_k (n+1)^{n+1-k} = \sum_{k=0}^n \binom{n+1}{k} B_k (n+1)^{n-k}$$

Following the "nonsense" above, we treat $B$ as an operator, $B^k=B_k$, and then the expression above becomes

$$\sum_{k=0}^{n} k^n = (B+n+1)^n$$

so that

$$\begin{aligned}&\lim_{n\to\infty} \frac{1^n + 2^n +\dots + n^n}{n^n}\\ =&\lim_{n\to\infty} \frac{(B+n+1)^n}{n^n}\\ =&\lim_{n\to\infty} \left(1+\frac{B+1}{n}\right)^n\\ =& e^{B+1} =e\times e^B\end{aligned}$$

Notice that

$$e^B=\sum_{k=0}^{\infty} \frac{1}{k!}B_k=\frac{1}{e-1}$$

is exactly the result given by the definition of the generating function of the Bernoulli numbers! Hence

$$\lim_{n\to\infty} \frac{1^n + 2^n +\dots + n^n}{n^n}=\frac{e}{e-1}$$

The Subtle Point

I still haven't fully grasped the subtlety of putting sequences into one-to-one correspondence with powers of an operator. I have a vague sense that it must be closely connected to the method of generating functions. If I make further progress on this, I'll be sure to share it with everyone.