IRL ABooklet

We will use the following lemma three times.
Lemma. For all $k \ge 1$ and all $1 \le M \le N$ we have
$$\sum_{n=M}^{N} \frac{1}{n(n+k)} < \frac{1}{k} \sum_{n=M}^{M+k-1} \frac{1}{n} .$$
Proof. Using $\frac{1}{n(n+k)} = \frac{1}{k} \left( \frac{1}{n} - \frac{1}{n+k} \right)$ we see that
$$\begin{align*} \sum_{n=M}^{N} \frac{1}{n(n+k)} &= \frac{1}{k} \sum_{n=M}^{N} \left( \frac{1}{n} - \frac{1}{n+k} \right) \\ &= \frac{1}{k} \sum_{n=M}^{M+k-1} \frac{1}{n} - \frac{1}{k} \sum_{n=N+1}^{N+k} \frac{1}{n} \\ &< \frac{1}{k} \sum_{n=M}^{M+k-1} \frac{1}{n} . \end{align*}$$
If $M \le N-k+1$ the terms $1/n$ for $M+k \le n \le N$ cancel out as they appear in both sums. When $N < M+k-1$, we have introduced extra terms which appear in both sums. In this case, the sum on the right hand side would only need to go up to $n=N$, but we do not need this stronger inequality. $\square$

We will use this lemma in two special cases:
$$k = 1 \qquad \sum_{n=M}^{N} \frac{1}{n(n+1)} < \frac{1}{M} \qquad (17)$$
$$M = 1 \qquad \sum_{n=1}^{N} \frac{1}{n(n+k)} < \frac{1}{k} \sum_{n=1}^{k} \frac{1}{n} \qquad (18)$$

Let now $S$ be a sum of a finite number of terms of the form $\frac{1}{mn(m+n+1)}$.
Let $N$ be such that no term with $n > N$ or with $m > N$ appears in $S$. For fixed $m$ we obtain from (18) with $k = m + 1$
$$\sum_{n=1}^{N} \frac{1}{mn(m+n+1)} < \frac{1}{m(m+1)} \sum_{n=1}^{m+1} \frac{1}{n} .$$
Hence
$$S < \sum_{m=1}^{N} \frac{1}{m(m+1)} \sum_{n=1}^{m+1} \frac{1}{n} .$$

Instead of summing by row, we can first add along the columns. This gives
$$\sum_{m=1}^{N} \frac{1}{m(m+1)} \sum_{n=1}^{m+1} \frac{1}{n} = \sum_{m=1}^{N} \frac{1}{m(m+1)} + \sum_{n=2}^{N+1} \frac{1}{n} \sum_{m=n-1}^{N} \frac{1}{m(m+1)}.$$
From (17) with $M = 1$ and with $M = n - 1$ we obtain
$$\sum_{m=1}^{N} \frac{1}{m(m+1)} < 1 \quad \text{and} \quad \sum_{m=n-1}^{N} \frac{1}{m(m+1)} < \frac{1}{n-1}.$$
Hence, $S < 1 + \sum_{n=2}^{N+1} \frac{1}{n(n-1)} = 1 + \sum_{n=1}^{N} \frac{1}{n(n+1)} < 2$ by (17) with $M = 1$.