It is easily checked that, for each positive integer $k$,$$\frac 1{3k-1}+\frac 1{3k}+\frac 1{3k+1}=\frac 1k+\frac 2{(3k)^3-3k}\tag1$$Set $k=1,2,\ldots,n$ in $(1)$ and add the $n$ equalities to find that$$\sum\limits_{k=2}^{3n+1}\frac 1k=\sum\limits_{k=1}^n\frac 1k+\sum\limits_{k=1}^n\frac 2{(3k)^3-3k}$$i.e$$\sum\limits_{k=1}^{2m+1}\frac 1{m+k}=1+\sum\limits_{k=1}^m\frac 2{(3k)^3-3k}\tag2$$The first three cases, $m=1,2,3,$ of $(2)$ are, respectively,$$\begin{align*}\frac 12+\frac 13+\frac 14 & =1+\frac 2{3^3-3}\\\frac 15+\frac 16+\cdots+\frac 1{13} & =1+\frac 2{3^3-3}+\frac 2{6^3-6}+\frac 2{9^3-9}+\frac 2{12^3-12}\\\frac 14+\frac 1{15}+\cdots+\frac 1{40} & =1+\frac 2{3^3-3}+\cdots+\frac 2{39^3-39}\end{align*}$$More generally, taking $m=1,2,\ldots,n$ in $(2)$ and adding the $n$ equalities, we find that$$\sum\limits_{k=1}^{(3^n-1)/2}\frac 1k=n+(n-1)\frac 2{3^3-3}+(n-2)\left(\frac 2{6^3-6}+\frac 2{9^3-9}+\frac 2{12^3-12}\right)+(n-3)\left(\frac 2{15^3-15}+\frac 2{18^3-18}+\cdots+\frac 2{39^3-39}\right)\tag3$$where there are $n$ expressions on the right-hand side of $(3)$.
Questions:
- How does setting $m=1,2,3$ into $(2)$ give the next three equations? I'm getting$$\begin{align*}\sum\limits_{k=1}^{3}\frac 1{k+1} & =\frac 12+\frac 13+\frac 14=1+\frac 2{(3k)^3-3k}\\\sum\limits_{k=1}^{5}\frac 1{k+2} & =\frac 13+\frac 14+\frac 15+\frac 16+\frac 17=1+\frac 2{3^3-3}+\frac 2{6^3-6}\\\sum\limits_{k=1}^7\frac 1{k+3} & =\frac 14+\frac 15+\cdots+\frac 1{10}=1+\frac 2{3^3-3}+\frac 2{6^3-6}+\frac 2{9^3-9}\end{align*}$$
- If $(2)$ is wrong, does that mean that the final equation $(3)$ is also wrong?
- How do you arrive at $(3)$?
I feel like the PDF is wrong. Whatever value of $m$ you set, the denominator of $1/(m+k)$ will always increase by one. So the third equation cannot exist. I also tried just ignoring the setting $m=1,2,3$ portion and just tried summing both sides for $1\leq m\leq n$ to get$$\sum\limits_{m=1}^n\sum\limits_{k=1}^{2m+1}\frac 1{m+k}=m+2\sum\limits_{m=1}^n\sum\limits_{k=1}^{2m+1}\frac 1{(3k)^3-3k}\tag4$$But I have never dealt with "nested summations" so far. Any tips on simplifying $(4)$?