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On page 1290 of Crandall 1978, there is a line saying

From Corollary (7.1) and Lemma (7.4) we have $$m(2^{A_k}-3^k)<\sum_{j=0}^{k-1}(3+\frac{1}{m})^{k-1}$$

But looking at Corollary (7.1) we have $$m(2^{A_k}-3^k)=\sum_{j=0}^{k-1}2^{A_j}3^{k-1-j}$$

and looking at Lemma (7.4) we have $$2^{A_j}\leq(3+\frac{1}{m})^{j}$$

So for me (7.1) and (7.4) would lead to $$m(2^{A_k}-3^k)<\sum_{j=0}^{k-1}(3+\frac{1}{m})^{j}3^{k-1-j}$$

but RHS is $$3^{k-1}\sum_{j=0}^{k-1}(1+\frac{1}{3m})^{j}=3^{k-1}\frac{(1+\frac{1}{3m})^{k}-1}{(1+\frac{1}{3m})-1}=m((3+\frac{1}{m})^{k}-3^k)$$

So it leads to $$m(2^{A_k}-3^k)<m((3+\frac{1}{m})^{k}-3^k)$$ $$2^{A_k}<(3+\frac{1}{m})^{k}$$ or (7.4) again

Where does the first line came from? Am I not looking at the right place in those Lemma?

user489810
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    It is just $3 < 3+\frac{1}{m}$. – Nicolas Jan 16 '21 at 19:55
  • oh!!! dumb of me. Thank you – user489810 Jan 16 '21 at 20:02
  • Btw, what is in the first (cited) formula? There is a sum-sign, but the sum-index $j$ is not used. So it seems as if this is a little obfuscation - because the sum is nothing else as the $k$-mutiple of the parenthese : $k \cdot (3+1/m)^{(k-1)} $ and when $k \ge 4$ we can even use $ (3+1/m)^k $ in an in-equality... Or did I miss some sophisticated reason for that expression? – Gottfried Helms Jan 18 '21 at 16:48
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    @Gottfried Helms, that's exactly what happens a few lines bellow ($k \cdot (3+1/m)^{(k-1)}$). I guess that's the whole point of replacing $3^{k-1-j}$ by $(3+\frac{1}{m})^{k-1-j}$ – user489810 Jan 18 '21 at 18:18

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