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Just finished Euler: The Master of us All. A good fraction of the book is dedicated to explaining in why certain divergent series were useful in proving Euler's theorems, but this one is never explained: \begin{align} 1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \cdots = 0.66215 + \frac{1}{2}\log(\infty)^{3} \end{align} I'm baffled by this expression. I'm ok with $\infty = \infty$, but why would it be reasonable to express it in this fashion, and what is the utility?

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    $\begingroup$ Are you sure of the power of $\log$? $\endgroup$ – Mhenni Benghorbal Jan 5 '15 at 3:22
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    $\begingroup$ @MhenniBenghorbal Does it matter at all?? $\endgroup$ – Timbuc Jan 5 '15 at 3:22
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    $\begingroup$ @Timbuc: off course it matters! $\endgroup$ – Mhenni Benghorbal Jan 5 '15 at 3:24
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    $\begingroup$ @MichaelHardy Listen carefully: I did'nt ask whether Euler thought of that as having a definite meaning or not, but whether Mhenni himself gives some definite meaning to that so thatt its powers matter. If you, or anyone else, don't know what Euler meant then I can't understand how that power may be important. Relying on authority, even Euler, for this issomething I can't do. $\endgroup$ – Timbuc Jan 5 '15 at 3:55
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    $\begingroup$ The ${}^3$ appears to be a footnote (rather poor style!); see the image in user157227's answer below with a ${}^4$ footnote shortly thereafter. $\endgroup$ – aes Jan 5 '15 at 4:53
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For $$ \left(\sum_{n=0}^{500} \frac 1 {2n+1}\right) - \frac 1 2\log(2\cdot500+1) = \left(\sum_{\text{odd }n\,\le\,1001} \frac 1 n\right) - \frac 1 2\log(1001) $$ I get $0.6368\ldots$. Might Euler have had in mind some sort of limit as the number in place of $500$ grows?

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    $\begingroup$ This sounds much likely, since we have $$ \sum_{k : \text{odd } \leq n} \frac{1}{k} = \frac{1}{2}\log n + \frac{\gamma+\log 2}{2} + o(1), $$ where the constant part has value $0.63518142\cdots$. $\endgroup$ – Sangchul Lee Jan 5 '15 at 5:13
  • $\begingroup$ Euler had the correct solution in mind, and the citation of him was wrong. And from this also the use of this formula to indicate Euler's fallability (one might use another case, but this case hasn't been a case of error) See my answer and my last comment to the OP's question. (Meanwhile I had mail-exchange with the author) $\endgroup$ – Gottfried Helms Jan 6 '15 at 19:48
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That is quite strange... since even it is understood as a precursory notation for asymptotic expansion, still we should get

$$ \sum_{k=0}^{n} \frac{1}{2k+1} = \frac{1}{2}\log n + C + \frac{1}{2n} - \frac{11}{48n^{2}} + \cdots, $$

where

$$ C = \frac{\gamma}{2} + \log 2 \approx 0.98175501301071173972. $$

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This expression is wrong and it is used to prove a point. Here is the first sentence in the paragraph above the expression.

The observation is that Euler was far from infallible.

The author is saying that even Euler makes mistakes.


Picture of page:

enter image description here

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  • $\begingroup$ The author goes on to state that some meaning can be found by the expressions. Namely that both sides of the first equation are infinite and that the second equation is correct when stated as a limit in the obvious way. $\endgroup$ – user157227 Jan 5 '15 at 3:25
  • $\begingroup$ *fall = far (typo that I'm not allowed to correct). $\endgroup$ – Reinderien Jan 5 '15 at 3:57
  • $\begingroup$ Quite funny... you forgot to scan the bottom of the page. The $3$ is actually the footnote number for reference, as the $4$ in the second formula. $\endgroup$ – Tom-Tom Jan 6 '15 at 13:59
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Possibly relevant:

$$\lim_{n\to\infty}1+\frac13+\frac15+\dotsb+\frac1n-\frac12\ln(n)\approx0.63518$$

The number's off, though…

Here's the full page from the book. Can anyone follow the footnote?

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  • $\begingroup$ please see one reference into Euler's original work in my new answer (Of course, having one reference it does not mean, that possibly Euler might have computed that value various times in various articles and one time erred with the computation/with the print. Don't have the source from the book which the OP refers to) $\endgroup$ – Gottfried Helms Jan 6 '15 at 14:04
  • $\begingroup$ I should mention that the constant is $\dfrac{\gamma+\ln2}2$. This follows from $\lim_{n\to\infty}1+\frac12+\frac13+\dotsb+\frac1n-\ln n=\gamma$ and from $1-\frac12+\frac13-\dotsb=\ln2$. $\endgroup$ – Akiva Weinberger Jan 6 '15 at 14:35
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After a little research in the Euler-book to which the author of "Euler: The master of us all" refers, it seems that the alleged wrong equation does not occur at all in Euler's book.
This is likely
a) because of failing textsearch for the number $0.66215$ as well as for $66215$ (other numbers and/or text can be found) and
b) because it does not seem that the book deals with that deeper number-theoretical problems at all.
Here is a searchable book-preview using google: the Hewlett-translation

So, the example of the author W. Dunham for the fallibility of L. Euler seems to be an unfounded, at least an unlucky one. In the contrary, the derivation of the formula actually occurs (for instance) in the english translation (Jordan Bell) of E47 ("Eneström-index") kept on arXiv and it contains also the correct value $= 0.6351814227307392$ . This occurs in item (30) on page 10 of the pdf-file.

Here is a picture of the scan of the original E047 at the Euler-archives (I've marked the number by a red box):
picture

Remark: Of course, having one reference does not mean that possibly Euler might have computed that value various times in various articles and one time erred with the computation/with the print. But it seems that it does not exist at all in the book, where the OP's literature points to

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