Showing $\gamma < \sqrt{1/3}$ without a computer

Question

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.

Answer 1

We may exploit an integral representation for $\gamma$ and some trivial bounds. Since: $$\begin{eqnarray*}\gamma&=&\lim_{n\to +\infty}(H_n-\log n) = \lim_{n\to +\infty}\left(\int_{0}^{1}\frac{x^n-1}{x-1}\,dx-\int_{1}^{n}\frac{dt}{t}\right)\\&=&\lim_{n\to +\infty}\left(\int_{0}^{+\infty}\frac{1-e^{-nx}}{e^x-1}\,dx-\int_{1}^{n}\int_{0}^{+\infty}e^{-tx}\,dx\,dt\right)\\&=&\lim_{n\to +\infty}\left(\int_{0}^{+\infty}\left(\frac{1}{1-e^{-x}}-\frac{1}{x}\right)e^{-x}\,dx+\int_{0}^{+\infty}\left(\frac{1}{x}-\frac{1}{e^x-1}\right)e^{-nx}\,dx\right)\end{eqnarray*}$$ but $$\frac{1}{x}-\frac{1}{e^x-1}\in\left[0,\frac{1}{2}\right]\quad\forall x\in\mathbb{R}^+,$$ the last integral is $O\left(\frac{1}{n}\right)$ and we simply have: $$\gamma = \int_{0}^{+\infty}\left(\frac{1}{1-e^{-x}}-\frac{1}{x}\right)e^{-x}\,dx.\tag{1}$$ Since $f(x)=\frac{1}{1-e^{-x}}-\frac{1}{x}$ is concave on $\mathbb{R}^+$, for any $x\in\mathbb{R}^+$ we have $f(x)\leq\frac{1}{2}+\frac{x}{12}$, so: $$\gamma\leq\int_{0}^{+\infty}\left(\frac{1}{2}+\frac{x}{12}\right)e^{-x}\,dx = \frac{7}{12}.$$ This bound is a bit worse than $\gamma\leq\frac{1}{\sqrt{3}}$, but it can be improved by considering further terms of the Taylor series of $f(x)$ around the origin.

Answer 2

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.