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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?

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

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