IRL_ABooklet_2023

We will use the following notation: $$H_i=\sum _{k=1}^i\frac{1}{k}{P}_i(m)=\sum _{k=m}^N\frac{H_{k+i}}{k(k+1)}.$$ Lemma 1. For all $i\ge 1$ and all $1\le M\le N$ we have $$\sum _{k=M}^N\frac{1}{k(k+i)}<\frac{1}{i}\sum _{k=M}^{M+i-1}\frac{1}{k}.$$ Proof. Using $\frac{1}{k(k+i)}=\frac{1}{i}\left(\frac{1}{k}-\frac{1}{k+i}\right)$ we see that $$\sum _{k=M}^N\frac{1}{k(k+i)}=\frac{1}{i}\sum _{k=M}^N\left(\frac{1}{k}-\frac{1}{k+i}\right)=\frac{1}{i}\sum _{k=M}^{M+i-1}\frac{1}{k}-\frac{1}{i}\sum _{k=N+1}^{N+i}\frac{1}{k}<\frac{1}{i}\sum _{k=M}^{M+i-1}\frac{1}{k}.$$ If $M+i-1\le N$ the terms $1/k$ for $M+i\le k\le N$ cancel out as they appear in both sums. When $M+i-1>N$, 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 $k=N$, but we don't need this stronger inequality. $\square$

$$i=2\sum _{k=M}^N\frac{1}{k(k+2)}<\frac{1}{2M}+\frac{1}{2(M+1)}(32)$$ $$M=1\sum _{k=1}^N\frac{1}{k(k+i)}<\frac{1}{i}\sum _{k=1}^i\frac{1}{k}=\frac{H_i}{i}(33)$$

Lemma 2. For all $N>m\ge 1$ and $i\ge 1$ we have $$P_i(m)<\frac{1}{m}{H}_{m+i}+\frac{1}{i}\sum _{k=m+1}^{m+i}\frac{1}{k}.$$ $$\begin{array}{rl}{P}_i(m)& =\sum _{k=m}^N\frac{H_{k+i}}{k(k+1)}\\ & =\sum _{k=m}^N\left(\frac{1}{k}-\frac{1}{k+1}\right)H_{k+i}=\sum _{k=m}^N\frac{1}{k}{H}_{k+i}-\sum _{k=m+1}^{N+1}\frac{1}{k}{H}_{k+i-1}\\ & =\frac{1}{m}{H}_{m+i}+\sum _{k=m+1}^N\frac{1}{k}(H_{k+i}-H_{k+i-1})-\frac{1}{N+1}{H}_{N+i}\\ & <\frac{1}{m}{H}_{m+i}+\sum _{k=m+1}^N\frac{1}{k(k+i)}<\frac{1}{m}{H}_{m+i}+\frac{1}{i}\sum _{k=m+1}^{m+i}\frac{1}{k}\text{using Lemma 1.}\end{array}$$ $\square$ In particular, we obtain $$P_1(m)<\frac{1}{m+1}+\frac{1}{m}{H}_{m+1}(34)$$ $$P_2(m)<\frac{1}{2(m+1)}+\frac{1}{2(m+2)}+\frac{1}{m}{H}_{m+2}.(35)$$

$$\begin{array}{rlrl}S& =\sum _{a,b,c=1}^N\frac{1}{abc(a+b+c+1)}=\sum _{a,b=1}^N\frac{1}{ab}\sum _{c=1}^N\frac{1}{c(a+b+c+1)}& & \\ & <\sum _{a,b=1}^N\frac{H_{a+b+1}}{ab(a+b+1)}& & \text{using (33) with }i=a+b+1\\ & =\sum _{a,b=1}^N\frac{a+b}{ab}\cdot \frac{H_{a+b+1}}{(a+b)(a+b+1)}& & \\ & =\sum _{k=2}^{2N}\sum _{a+b=k}\left(\frac{1}{a}+\frac{1}{b}\right)\frac{H_{k+1}}{k(k+1)}=2\sum _{k=2}^{2N}\sum _{m=1}^{k-1}\frac{1}{m}\cdot \frac{H_{k+1}}{k(k+1)}.& & \end{array}$$ \sum_{k=2}^{2N} \sum_{m=1}^{k-1} \frac{1}{m} \cdot \frac{H_{k+1}}{k(k+1)} = \sum_{m=1}^{2N-1} \frac{1}{m} \sum_{k=m+1}^{2N} \frac{H_{k+1}}{k(k+1)}. $$Therefore,$$ $$\begin{array}{rlrl}S<2\sum _{m=1}^{2N-1}\frac{1}{m}\sum _{k=m+1}^{2N}\frac{H_{k+1}}{k(k+1)}& =2\sum _{m=1}^{2N-1}\frac{1}{m}{P}_1(m+1)& & \\ & <2\sum _{m=1}^{2N-1}\frac{1}{m}\left(\frac{1}{m+2}+\frac{1}{m+1}{H}_{m+2}\right)& & \text{using (34)}\\ & =2\sum _{m=1}^{2N-1}\frac{1}{m(m+2)}+2\sum _{m=1}^{2N-1}\frac{H_{m+2}}{m(m+1)}& & \\ & =2\sum _{m=1}^{2N-1}\frac{1}{m(m+2)}+2P_2(1)& & \\ & <1+\frac{1}{2}+2\left(\frac{1}{4}+\frac{1}{6}+H_3\right)=6.& & \text{using (32) and (35).}\end{array}$$ $$