TITLE: Showing $\gamma < \sqrt{1/3}$ without a computer
QUESTION [14 upvotes]: In 1735 Euler gave the value of $\gamma$ as $0.577218.$ The constant is generally defined as the limit of the difference between the harmonic series and $\log n:~\gamma= \lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{k}-\log n.$ Euler apparently relished this sort of calculation and must have taken quite a few terms to get such a good approximation.
My question is whether without a computer one can now prove that
$$\gamma <  \sqrt{1/3}$$ 
with at least some savings in terms of the type of work Euler apparently expended? I don't think there's any point in raising the bar to $\gamma < \ln 2\sqrt{\frac{23}{29}} $  because it seems to be the same sort of question.     
My own thought was to compare $\frac{1}{2}\int_0^{1/3}\frac{dx}{\sqrt{x}}$ to something like $(-1)\cdot\int_0^\infty e^{-u}\ln u~du$ but I expect there's a better way.

REPLY [2 votes]: Well, you need a hand calculator but only to get $\ln(3)$ or $\ln(4)$.
Using Euler's summation formula for $H_n = \sum(1/n)$, and then isolating $\gamma$ on the left side, you have:
$$
\gamma = H_n - \ln (n) -\frac{1}{2n}+\frac{1}{12n^2} - \frac{1}{120n^4} + \frac{\theta}{256n^6}$$
where $0 < \theta < 1$.  
One would think you need to go out a ways in the harmonic series to get an excellent approximation to $\gamma$, but with these three asymptotic terms you need only go to $n=3$.
We get $0.5772197 < \gamma < .5772162 < .57735 < \frac{1}{\sqrt{3}}$  for $n=3$.  
For $n=4$ we need only to the $n^2$ term to determine that
$$ \gamma < 0.577247$$
and going to the $n^4$ accuracy we have here, we can see that 
$$
0.57721475 < \gamma < 0,57721573 < \frac{1}{\sqrt{3}}
$$
And yes, unless my hand arithmetic is wrong, you have an error in your 6-th decimal place of Euler's constant, which should be 5 or 6, not 8.