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 $$\begin{array}{rl}\sum _{n=M}^N\frac{1}{n(n+k)}& =\frac{1}{k}\sum _{n=M}^N\left(\frac{1}{n}-\frac{1}{n+k}\right)\\ & =\frac{1}{k}\sum _{n=M}^{M+k-1}\frac{1}{n}-\frac{1}{k}\sum _{n=N+1}^{N+k}\frac{1}{n}\\ & <\frac{1}{k}\sum _{n=M}^{M+k-1}\frac{1}{n}.\end{array}$$ 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: $$k=1\sum _{n=M}^N\frac{1}{n(n+1)}<\frac{1}{M}(17)$$ $$M=1\sum _{n=1}^N\frac{1}{n(n+k)}<\frac{1}{k}\sum _{n=1}^k\frac{1}{n}(18)$$

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 $$\sum _{m=1}^N\frac{1}{m(m+1)}<1\text{and}\sum _{m=n-1}^N\frac{1}{m(m+1)}<\frac{1}{n-1}.$$ 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 $M=1$.