$\sum_{n=0}^{\infty} \frac{1}{2n+1} = 0.66215 + \frac{1}{2}\log(\infty)^{3}$ 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?
 A: 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 
$$ C = \frac{\gamma}{2} + \log 2 \approx 0.98175501301071173972. $$
A: 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:

A: 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?
A: 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?
A: 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):
 
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 
