If $f(r)=1+\frac 12 +\frac 13+..+\frac 1r$ and $\sum_{r=1}^n (2r+1)f(r)=P(n)f(n+1)-Q(n)$, where $P$ and $Q$ are polynomial functions.

Question

If $$f(r)=1+\frac 12 +\frac 13+..+\frac 1r$$ and $$\sum_{r=1}^n (2r+1)f(r)=P(n)f(n+1)-Q(n)$$ where $P$ and $Q$ are polynomial functions,

Prove that

1. $$\sum_{r=0} ^{10} P(r)=506$$
2. $$\sum_{r=0}^{\infty} \frac{1}{Q(r)} = 2$$
3. $$P(13) -Q(13) = 91$$

For the expression inside the sigma function $$(2r+1)(f(r))$$ $$=((r+1)^2-r^2)f(r)$$ $$=(r+1)^2 (f(r)-f(r+1))+(r+1)^2 f(r+1) -r^2f(r)$$ $$=(r+1)^2 (-\frac{1}{r+1}) + (1+\frac 12 +\frac 13...\frac 1r)(2r+1) +(r+1)^2 \frac{1}{r+1}$$ $$(2r+1)f(r)$$

This was just an attempt to solve it, as you can see I just circled back to the original function. How should I solve it?

Answer 1

$$f(r)=1+\frac{1}{2} +\frac{1}{3}+..+\frac{1}{r}$$ is known as Harmonic series. It is easy to see the telescoping property $$f(r+1)-f(r)=\frac{1}{r+1}$$

Proceeding from $$ \begin{align*} & (2^2-1^2)f(1) + (3^2-2^2)f(2) + (4^2-3^2)f(3) + \ldots + ((n+1)^2-n^2)f(n) \\ & =-1^2f(1)-2^2(f(2)-f(1))-3^2(f(4)-f(3)) + \ldots + (n+1)^2f(n) \\ & =-1^2-\frac{2^2}{2}-\frac{3^2}{3} + \ldots + (n+1)^2f(n) \\ & =(n+1)^2\color{blue}{f(n)}-(1+2+ \ldots +n) \\ & =(n+1)^2\color{blue}{(f(n+1)-\frac{1}{n+1})}-(1+2+ \ldots +n) \\ & =(n+1)^2f(n+1)-(1+2+ \ldots +(n+1)) \\ & =(n+1)^2f(n+1)-\frac{(n+1)(n+2)}{2} \end{align*} $$

So we get $$P(n)=(n+1)^2$$ and $$Q(n)=\frac{(n+1)(n+2)}{2}$$

Can you finish the rest?