Why is $1+\sum\limits_{k=0}^{j-1}\frac{1\cdot2\cdot3\dots k}{3\cdot4\cdot\dots k+2}=3-\frac{2}{j+1}$

Question

Why is $1+\sum\limits_{k=0}^{j-1}\frac{1\cdot2\cdot3\dots k}{3\cdot4\cdot\dots k+2}=3-\frac{2}{j+1}$

If one has the result it is not difficult to verify it by induction, but how can I solve it without induction ?

Answer 1

HINT:

$$\prod_{r=1}^n\dfrac r{r+2}=\dfrac{1\cdot2}{(n+1)(n+2)}=2\cdot\dfrac{n+2-(n+1)}{(n+1)(n+2)}=?$$

See Telescoping Series

Answer 2

We have : $$ \begin{split} \require{cancel} \dfrac{1\cdot2\cdot\bcancel{3\cdot4\cdot5\cdots k}}{\bcancel{3\cdot4\cdot5\cdots k}\cdot k+1\cdot k+2}&=\dfrac{1\cdot2}{k+1\cdot k+2}\\&=\dfrac{2}{k+1}-\dfrac{2}{k+2} \end{split} $$

So the summation $S_j=\sum_{k=0}^{j-1}\dfrac{1\cdot2\cdots k}{3\cdot4\cdots k+2}$ is a telescoping sum and is equal to:

$$ \begin{split} S_j&=\sum_{k=0}^{j-1}\left(\dfrac{2}{k+1}-\dfrac{2}{k+2}\right) \\&=2\left(1-\dfrac{1}{j+1}\right) \end{split} $$

Answer 3

We have

$$\begin{align} 1+\sum_{k=1}^{j-1}\frac{1\cdot 2 \cdot 3\cdots k}{3\cdot 4\cdot5\cdots k+2}&=1+\sum_{k=0}^{j-1}\frac{k!}{(k+2)!/2}\\\\ &=1+2\sum_{k=0}^{j-1}\frac{1}{(k+1)(k+2)}\\\\ &=1+2\sum_{k=0}^{j-1}\left(\frac{1}{k+1}-\frac{1}{k+2}\right)\\\\ &=1+2\left(1-\frac{1}{j+1}\right)\\\\ &=3-\frac{2}{j+1} \end{align}$$