How to do this?

Question

If $$\sum_{r=1}^n t_r = n(n+1)(n+2)(n+3)$$ then find out

$$\ \lim_{n\to\infty} \sum_{r=1}^n \frac{1}{t_r}=?$$

Answer 1

Hints:

$$t_n=n (n + 1) (n + 2) (n + 3) - (n-1) n (n + 1) (n + 2) = 4 n (n + 1) (n + 2) $$

so decompose into partial fractions

$$\frac{1}{ 4 n (n + 1) (n + 2)} = \frac18\left(-\frac{1}{n} + \frac{1}{n + 1}+ \frac{1}{n + 1}- \frac{1}{n + 2}\right) $$

then telescope the sum

Answer 2

First, you need to solve for $t_n$: $$t_n = \sum_{r = 1}^{n}t_r - \sum_{r = 1}^{n-1}t_r = n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2) = 4n(n+1)(n+2)$$

After this, you can use partial fractions and then simplify the telescoping sum $$\sum_{r = 1}^{n}\dfrac{1}{t_r} = \sum_{r = 1}^{n}\dfrac{1}{4r(r+1)(r+2)} = \sum_{r = 1}^{n}\left[\dfrac{\tfrac{1}{8}}{r} - \dfrac{\tfrac{1}{4}}{r+1} + \dfrac{\tfrac{1}{8}}{r+2}\right] = \cdots$$