How to proceed to show this hold by induction?

Question

Show that $$\frac3{1\cdot2\cdot4}+\frac4{2\cdot3\cdot5}+\dots+\frac{n+2}{n(n+1)(n+3)}=\frac16\left[\frac{29}6-\frac4{n+1}-\frac1{n+2}-\frac1{n+3}\right],\text{ for $n\in\mathbb N$}.$$

I try induction. For $n=1$, it is trivial and then let it is true for some $k\in\mathbb N.$ Then for $k+1$,we have to show $$\frac3{1\cdot2\cdot4}+\frac4{2\cdot3\cdot5}+\dots+\frac{k+3}{(k+1)(k+2)(k+4)}=\frac16\left[\frac{29}6-\frac4{k+2}-\frac1{k+3}-\frac1{k+4}\right],$$ which is quivalent to show $$\frac16\left[\frac{29}6-\frac4{k+1}-\frac1{k+2}-\frac1{k+3}\right]+\frac{k+3}{(k+1)(k+2)(k+4)}=\frac16\left[\frac{29}6-\frac4{k+2}-\frac1{k+3}-\frac1{k+4}\right]$$ by induction hypothesis. Now is there any easier trick other than long and tedious computation?

Several terms cancel obviously, and the remaining terms look simple. Why not continue as you have done. — Macavity, Dec 01 '13 at 09:17
Do you consider partial fraction decomposition to be long and tedious? — David H, Dec 01 '13 at 09:18
This is equivalent to showing $$\frac{k+3}{(k+1)(k+2)(k+4)} = \frac16\left[\frac4{k+1}-\frac3{k+2}-\frac1{k+4}\right]$$ Now that really doesn't look tough. — Macavity, Dec 01 '13 at 09:20
Just set off similar terms from the RHS and the first part of the LHS. — Macavity, Dec 01 '13 at 09:22

score 1 · Accepted Answer · answered Jun 01 '16 at 18:21

$$\begin{eqnarray*}\frac{n+2}{n(n+1)(n+3)}&=&\frac{1}{n(n+1)}\left(1-\frac{1}{n+3}\right)\\&=&\left(\frac{1}{n}-\frac{1}{n+1}\right)\left(1-\frac{1}{n+3}\right)\\&=&\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n(n+3)}+\frac{1}{(n+1)(n+3)}\\&=&\frac{1}{n}-\frac{1}{n+1}-\frac{1}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)+\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n+3}\right)\\&=&\frac{\frac{2}{3}}{n}-\frac{\frac{1}{2}}{n+1}-\frac{\frac{1}{6}}{n+3}\end{eqnarray*}$$ (that can be done through the residue theorem, too), hence, by setting $H_n=\sum_{k=1}^{n}\frac{1}{k}$: $$\begin{eqnarray*} \sum_{n=1}^{N}\frac{n+2}{n(n+1)(n+3)}&=&\frac{2}{3}\sum_{n=1}^{N}\frac{1}{n}-\frac{1}{2}\sum_{n=1}^{N}\frac{1}{n+1}-\frac{1}{6}\sum_{n=1}^{N}\frac{1}{n+3}\\&=&\frac{2}{3}\sum_{n=1}^{N}\frac{1}{n}-\frac{1}{2}\sum_{n=2}^{N+1}\frac{1}{n}-\frac{1}{6}\sum_{n=4}^{N+3}\frac{1}{n}\\&=&\left(\frac{2}{3}-\frac{1}{2}-\frac{1}{6}\right)H_N-\frac{1}{2}\left(\frac{1}{N+1}-H_1\right)-\frac{1}{6}\left(\frac{1}{N+1}+\frac{1}{N+2}+\frac{1}{N+3}-H_3\right)\\&=&\frac{H_1}{2}+\frac{H_3}{6}-\frac{1}{6}\left(\frac{4}{N+1}+\frac{1}{N+2}+\frac{1}{N+3}\right)\\&=&\color{red}{\frac{29}{36}-\frac{1}{6}\left(\frac{4}{N+1}+\frac{1}{N+2}+\frac{1}{N+3}\right)} \end{eqnarray*}$$

as wanted.

Thank you very much.(Nice 'hat' in profile pic by the way :)) — Silent, Jun 05 '16 at 13:50

How to proceed to show this hold by induction?

1 Answers1