# Did I find a mistake in a classic proof? If not, could someone help me understand?

I am working through Denis Hanson's proof that LCM$$(1,2,3,\dots,n) < 3^n$$

To be clear, the mistake is minor and does not affect the result achieved. Still, I was surprised.

Here's what I believe is wrong (on page 36):

Observe that $$\frac{a_1-1}{a_1} + \frac{a_2-1}{a_2} + \dots + \frac{a_k-1}{a_k} = \left(1 - \frac{1}{a_1}\right) + \dots + \left(1 - \frac{1}{a_k}\right)$$ $$= k - 1 + \frac{1}{a_{k+1}+1}$$

where $$a_i$$ is defined as:

• $$a_1 = 2$$
• $$a_{n+1} = a_1 a_2 \dots a_n + 1$$

I believe that the corrected version should read:

$$= k - 1 + \frac{1}{a_{k+1} - 1}$$

Here is my reasoning:

• It is true for $$k=1$$ since $$\frac{a_1 -1 }{a_1} = \frac{2-1}{2} = (1 - 1) + \frac{1}{3 - 1} = (1 - 1) + \frac{1}{a_2 - 1}$$
• Assume that my hypothesis is true up to some $$k \ge 1$$ $$\frac{a_1 - 1}{a_1} + \dots + \frac{a_{k} - 1}{a_k} = k - 1 + \frac{1}{a_{k+1}-1}$$
• $$\frac{a_1 - 1}{a_1} + \dots + \frac{a_{k+1} - 1}{a_{k+1}} = (k - 1 + \frac{1}{a_{k+1}-1}) + 1 - \frac{1}{a_{k+1}}=$$ $$(k+1) - 1 + \frac{a_{k+1}}{a_{k+1}(a_{k+1}-1)} - \frac{a_{k+1}-1}{a_{k+1}(a_{k+1}-1)} = (k+1) - 1 + \frac{1}{a_{k+1}^2 - a_{k+1}} =$$ $$(k+1) - 1 + \frac{1}{a_{k+2}-1}$$

Am I wrong?

• I think you are right. (You have a typo in the last bullet point. $(a_{k+1}-1)/a_k$ should be $(a_{k+1}-1)/a_{k+1}$). – mathlove Oct 10 '18 at 16:34
• Thanks! I fixed the typo. I appreciate all your help! – Larry Freeman Oct 10 '18 at 16:43