If $ \sum_{r=1}^{n} t_r $ = $ { n ( n + 1 ) ( n + 2 ) ( n + 3 ) } \over {8} $ , then what does $ \sum_{r=1}^{n} {1\over {t_r} } $ equal to?

Question

$ \sum_{r=1}^{n} t_r $ = $ { n ( n + 1 ) ( n + 2 ) ( n + 3 ) } \over {8} $

$ \sum_{r=1}^{n} {1\over {t_r} } $ = ?

My progress so far :

$ t_1 = {{1*2*3*4 } \over 8 } = 3 $

$ t_2 = 15 $

$ t_3 = 45 $

$ t_4 = 105 $

$ t_5 = 210 $

$ \sum_{r=1}^{n} {1\over {t_r} } $ = 1/3 + 1/12 + 1/30 + 1/60 + 1/105 .... So on , but I dont get a pattern .

How do I solve this ?

Answer 1

You can verify the series simplification at the bottom via telescoping or induction.

$$\sum_{r=1}^{n}t_r=\frac{n(n+1)(n+2)(n+3)}{8}$$ $$\implies t_n=\sum_{r=1}^{n}t_r-\sum_{r=1}^{n-1}t_r=\frac{n(n+1)(n+2)(n+3)}{8}-\frac{(n-1)n(n+1)(n+2)}{8}=\frac{n(n+1)(n+2)}{8}\big((n+3)-(n-1)\big)=\frac{n(n+1)(n+2)}{2}$$ $$\implies \frac{1}{t_r}=\frac{2}{r(r+1)(r+2)}=\frac{1}{r}-\frac{2}{r+1}+\frac{1}{r+2}$$ $$\implies \sum_{r=1}^n\frac{1}{t_r}=\sum_{r=1}^n\frac{1}{r}-2\sum_{r=1}^n\frac{1}{r+1}+\sum_{r=1}^n\frac{1}{r+2}=\frac{n(n+3)}{2(n+1)(n+2)}$$