A hand waving approach to a problem posted by Cut the Knot

Saw this tweet yesterday:

It was a fun problem to think about and the two solutions on the site use the Stolz-Cesaro Lemma, which is basically l’Hospital’s rule for sums.

Through the various Christmas preparations yesterday I was wondering if there was a simple way to see why the limit exists in the first place. What follows below isn’t a rigorous proof (or even close to one!) but instead how I convinced myself that the limit probably does exist.

Since seeing Tim Gowers “live blog” his solution to an old IMO problem, I’ve been interested in occasionally sharing the solution process rather than polished solutions to problems. Two examples of problems I’ve used for that idea are below:

A Challenge / Plea to math folks

A challenge relating to a few problems giving my son trouble

So, for the problem at hand, here’s my “hand waving” approach to convincing myself that the limit even existed:

We know that:

\lim_{x\to\infty} E_n = \lim_{x\to\infty} (1 + \frac{1}{n})^n = e

and that

\lim_{x\to\infty} H_n = \lim_{x\to\infty} (1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n}) \approx \ln{n} + \gamma

So I’ll approximate the difference we are considering like this for large n:

(E_{n+1})^{H_{n+1}} - (E_n)^{H_n}

\approx e^{H_{n+1}} - e^{H_n}

\approx (e^{H_n})(e^{\frac{1}{n+1}} - 1)

\approx (e^{\ln{n} + \gamma}) (\frac{1}{n+1} + \frac{1}{2(x+1)^2} + \ldots )

\approx (e^{\gamma}) * n * (\frac{1}{n+1} + \frac{1}{2(n+1)^2} + \ldots )

\approx e^{\gamma}

The fact that this hand waving approach arrives at the “right” answer is just a coincidence as I’m playing pretty fast and loose with limit rules. But, at least I now have some indication that this strange (and lovely!) \infty - \infty limit might actually exist.

Just for fun here’s what the expression looks like for n up to 1,000,000:


Definitely a fun little problem to noodle over 🙂



4 Comments so far. Leave a comment below.
  1. tomas,

    Hi, how come, that last paren gives 1/n please?

    • Because e^x = 1 + x + x^2 / 2 + x^3 / 6 + . . . .

      So, (I’ll just use 1/n instead of 1/(n+1) for simplicity,

      e^(1/n) = 1 + (1/n) +(1/2)* (1/n)^2 + (1/6)*(1/n)^3 + . . . .

      • tomas,

        I thought this: (1/(n+1) + 1/2*1/(n+1)^2+…) should give 1/n for it to cancel with n in penultimate row of your proof, right? But why?

      • Oh, sorry, I misunderstood and maybe should have included one more line in my sketch.

        The “n” multiplies through and leaves terms that look like 1 + 1/(2n) and smaller terms. As n goes to infinity the terms of order 1/n and smaller go to zero.

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