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Please consider the following self-contained excerpt from Chapter $1$ of Littlewood's A MATHEMATICIAN'S MISCELLANY. I have two questions:

1) How is the second (weak) inequality derived?

2) How does the result follow from the two (weak) inequalities?

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ryang
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  • Thomas has satisfactorily answered Question 2 above, so Question 1 has been reposted at http://math.stackexchange.com/q/417551/21813 – ryang Jun 11 '13 at 15:25

1 Answers1

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I don't think the part starting with "Suppose..." is related to this proof, but rather Littlewood is going to the next topic.

The integral formula for the area, plus the inequality $OP^2+OQ^2\leq 1$ is enough to deduce that the area is at most $\frac{\pi}{4}$.

I don't know why the inequality is true, but if $x>0$ and $a_n=xn$ then $$\frac{1+a_{n+1}}{a_n} = 1+ \frac{1+x}{x}\frac{1}{n}$$

So $$\lim_{n\to\infty}\left(\frac{1+a_{n+1}}{a_n}\right)^n = e^{\frac{1+x}{x}}$$

So the $\limsup$ above can be made arbitrarily close to $e$, so $e$ is the best lower bound.

Thomas Andrews
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  • Oh yes, of course, thank you. I was (and am still) stuck on my first (primary) question, and hadn't pondered hard enough over my second. Any idea why the given lim sup is at least $e$? Now I'm also wondering why the result is "best possible"-- best possible when compared against what? – ryang Jun 11 '13 at 14:46
  • Assuming that inequality, I can explain why $e$ is the "best possible" lower bound for the inequality, but I don't know why the inequality is true... Is there a footnote referenced to the asterisk? – Thomas Andrews Jun 11 '13 at 14:58
  • Hey Thomas, thanks for your clear explanation of the "best possible" part! Littlewood actually uses the bracketing asterisks to signal to the general reader technical passages (though evidently each passage may contain unrelated ideas). From what I see, the only reference is to George Polya (as you can see in the text). – ryang Jun 11 '13 at 15:15
  • You might get a proof of the inequality here if you just ask with the "Suppose..." sentence. The geometry problem confused the issue. – Thomas Andrews Jun 11 '13 at 15:22
  • Yes, just did: http://math.stackexchange.com/q/417551/21813 – ryang Jun 11 '13 at 15:25
  • Ideally, you'd find an example such that $\limsup$ is exactly $e$, otherwise we haven't shown that $\limsup = e$ is possible. – Thomas Andrews Jun 11 '13 at 15:47