Showing that $\frac{1}{18} + \frac{1}{19} + \cdots + \frac{1}{47} < 1$ without brute force calculation

Because $1/x$ is a convex function, there is an inequality $1/x\leq \int_{x-1/2}^{x+1/2} t^{-1}\,dt$. It follows that $$ \frac{1}{18}+\frac{1}{19}+\ldots+\frac{1}{47}\leq \int_{17.5}^{47.5}\frac{dt}{t}=\ln\left(\frac{47.5}{17.5}\right). $$ By hand, we can check that $2.715\cdot 17.5=47.5125>47.5$, so that $\frac{47.5}{17.5}<2.715<e$, and therefore $$ \ln\left(\frac{47.5}{17.5}\right)<1. $$

In here, D.W. DeTemple established that the sequence

$$R_n = 1+\frac{1}{2}+\frac{1}{3} + \ldots + \frac{1}{n}-\ln\left(n+\frac{1}{2}\right)$$

is strictly decreasing and converges very fast toward the Euler-Mascheroni constant $\gamma$. Notice that the harmonic series

$$H_n = R_n + \ln\left(n+\frac{1}{2}\right)$$

So

$$R_{47} < R_{17} \Rightarrow H_{47}-\ln\left(47+\frac{1}{2}\right) < H_{17}-\ln\left(17+\frac{1}{2}\right)$$

Therefore

$$H_{47} - H_{17} < \ln\left(\frac{95}{2}\right)-\ln\left(\frac{35}{2}\right) < 1$$

The last inequality is equivalent with $95 < 35e$, which is true and can be verified by hand if we approximate $e$ to three decimals.

Notice that the same approach does not work with the classic definition

$$\gamma = \lim_{n \to \infty} \left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}-\ln n\right)$$

because $H_n - \ln n$ converges very slow toward $\gamma$.

The integer numbers $$n_k:=\biggl\lceil{10000\over k}\biggr\rceil\geq{10^4\over k}$$ are easy to calculate by hand for the thirty numbers $k\in[18\,..\,47]$. Since $$\sum_{k=18}^{47}\>n_k=9999<10^4$$ we arrive at the desired estimate.