Is there a simple way to show this bound for Euler’s constant:
$$ \frac{1}{2n+2} < \sum_{k=1}^n\frac{1}{k} - \ln(n) - \gamma < \frac{1}{2n}$$
For proving convergence, I showed that $\sum_{k=1}^n 1/k - \ln(n)$ is bounded and monotonic, using the property
$$\frac{1}{k+1} < \int_k^{k+1} x^{-1} dx <\frac{1}{k}$$
This showed $$\frac{1}{n} < \sum_{k=1}^n \frac{1}{k} - \ln(n) < 1$$
But this does not help prove the error bounds.
Thanks
Use convexity.
Since $f(x) = 1/x$ is both decreasing and convex we have for $k \leqslant x \leqslant k+1$,
$$\frac{f(k+1) - f(k+2)}{k + 2 - (k+1)} \leqslant \frac{f(x) - f(k+1)}{k + 1 - x} \leqslant \frac{f(k) - f(k+1)}{k + 1 - k} $$
Hence,
$$\tag{*}\left(\frac{1}{k+1} - \frac{1}{k+2}\right)(k + 1 - x) \leqslant \frac{1}{x}- \frac{1}{k+1} \leqslant \left(\frac{1}{k} - \frac{1}{k+1} \right) (k+1 - x)$$
Note that
$$\int_{k}^{k+1}(k + 1 - x) \, dx = \left.-\frac{1}{2} (k+1 -x)^2\right|_k^{k+1} = \frac{1}{2}$$
Integrating (*) over $[k,k+1]$ and summing from $k = n$ to $N-1$ we get
$$\frac{1}{2(n+1)} - \frac{1}{2(N+1)}\leqslant \sum_{k=n}^{N-1} \int_{k}^{k+1} \frac{dx}{x}- \sum_{k=n}^{N-1} \frac{1}{k+1} \leqslant \frac{1}{2n} - \frac{1}{2N},$$
which reduces to
$$\frac{1}{2(n+1)} - \frac{1}{2(N+1)}\leqslant \sum_{k=1}^n\frac{1}{k} -\log n - \left(\sum_{k=1}^N \frac{1}{k}- \log N\right) \leqslant \frac{1}{2n} - \frac{1}{2N}$$
Then take the limit as $N \to \infty$.