sequences and series - Can one show that $sum_{n=1}^Nfrac{1}{n} -log N - gamma leqslant frac{1}{2N}$ without using the Euler-Maclaurin formula?

I would like to prove that

$$\sum_{n=1}^N\frac{1}{n} -\log N - \gamma \leqslant \frac{1}{2N}$$
without using the Euler-Maclaurin summation formula. The motivation for this is that I have come very close to doing so (see the answer provided below) but annoyingly have not actually proved the above.

Some may ask why I don't just use the formula. I'm writing a set of analytic number theory notes for my own use and it seems an unwieldy result to introduce and prove, given that the above inequality is all I need, and given that I have gotten so close without using Euler-Maclaurin!

Answer

Let
$$\gamma_n = \sum_{k=1}^n \frac{1}{k} - \log n.$$
Our goal is to show that

$$\gamma_n - \lim_{m \to \infty} \gamma_m \leq \frac{1}{2n}.$$
It is enough to show that, for $n $$\gamma_n - \gamma_m \leq \frac{1}{2n}.$$
This has the advantage of dealing solely with finite quantities.

Now,
$$\gamma_n - \gamma_m = \int_{n}^m \frac{dt}{t} - \sum_{k=n+1}^m \frac{1}{k} =\sum_{j=n}^{m-1} \int_{j}^{j+1} \left( \frac{1}{t} - \frac{1}{j+1} \right) \cdot dt .$$

At this point, if I were at a chalkboard rather than a keyboard, I would draw a picture. Draw the hyperbola $y=1/x$ and mark off the interval between $x=n$ and $x=m$. Divide this into $m-n$ vertical bars of width $1$. Each bar stretches up to touch the hyperbola at its right corner. There is a little wedge, bounded by $x=j$, $y=1/(j+1)$ and $y=1/x$. We are adding up the area of each of these wedges.¹

Because $y=1/x$ is convex, the area of this wedge is less than that of the right triangle with vertices at $(j,1/(j+1))$, $(j+1, 1/(j+1))$ and $(j,1/j)$. This triangle has base $1$ and height $1/j - 1/(j+1)$, so its area is $(1/2) (1/j - 1/(j+1))$. So the quantity of interest is
$$\leq \sum_{j=n}^{m-1} \frac{1}{2} \left( \frac{1}{j} - \frac{1}{j+1} \right) = \frac{1}{2} \left( \frac{1}{n} - \frac{1}{m} \right) \leq \frac{1}{2n}.$$

Of course, this is just a standard proof of Euler-Maclaurin summation, but it is a lot more geometric and easy to follow in this special case.

¹ By the way, since this area is positive, we also get the corollary that $\gamma_n - \gamma_m > 0$, so $\gamma_n - \gamma >0$, another useful bound.