Speaking of limits, here is a cool one!

So, if I ever make a list with the ten most counter-intuitive things (at least, at first glance), remind me to put into number 4 this limit :

\lim_{n\to\infty}n\sin(2\pi\cdot e\cdot n!)

I won’t lie. The first time that I looked at this beast, a small voice inside my head was saying “does not exist”.  The reason was that because of the sine and it’s periodicity, I was waiting for the frequency with which it oscillates to simply screw things up in the same way that the limit

\lim_{n\to\infty}n\sin(n\cdot e)

doesn’t exist.

I was also aware of limits like \lim_{n\to\infty}\frac{\sin(1/n)}{1/n}=\lim_{n\to\infty}n\sin(1/n)=1 but I didn’t make the connection at the time.

So you can imagine how I was when I found out that this limit does exist and is equal to 2π. Since I didn’t have a photographer near, this image will suffice.

I will give you links with the proof at the end. For me, the important part is to have first a good intuition on why this works like that.

So, let’s forget at this point about 2π and focus on the e\cdot n!  part. Let’s also keep in mind that e=\sum_{k=0}^{\infty}\frac{1}{k!} and that we can break that sum into two parts. By multiplying both by n!, we get a sum for the integer part and one for the fractional part of n!\cdot e.

Let’s see what happens!

Supposing that I_{n} is the sequence of integer parts of n! e and F_{n} is the sequence of fractional parts, we get

\sin(2\pi n!e)=\sin(2\pi I_{n} + 2\pi F_{n})=\sin(2\pi F_{n})

Thus, we just need to find the limit of F_{n} and, hopefully, we will be done. What should be the limit of F_{n}=n!e-[n!e]?

Returning to the series, we can write that fractional part as


from which we can see nice and well that F_{n}\to 0 while n\to\infty (try to bound it).

Thus, \sin(2\pi F_{n})\to 0 and, by using the idea mentioned earlier about n\sin(1/n), we can work our way to the full result.

Here’s a link to the proof. Kudos to the Calculus Humor guys for this nice problem.


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