How to find $\sum_{r=1}^{n}\frac{1}{T_r}$ when $\sum_{r=1}^{n}T_r$ is given?

Question

Question:

If $$\sum_{r=1}^{n}T_r=\frac{n}{8}(n+1)(n+2)(n+3)$$

Then find

$$\sum_{r=1}^{n}\frac{1}{T_r}$$

My Attempt:

I tried to find the terms $T_r$ by $S_r-S_{r-1}$ for $r>2$. For $r=1$, its simply $S_1$. Here $S_r$ represents sum till $r$ terms. I did that to find if I could see any properties relating the terms. As far as I did, I didn't note any kind of relation. So I gave up the process of finding the terms. I think, I need to find more terms in order to obtain a conclusion, but I felt that must a very length process.

So, I wish to know how to solve this problem. Is there any algorithm like replace $n$ by $\frac{1}{n}$ or anything like that to evaluate these kind of problems. The replacement that I suggested doesn't work for this problem.

Kindly guide me in this regard.

Answer 1

First we find $T_n$: $$ T_n = S_n - S_{n-1} = \frac{n(n+1)(n+2)(n+3)}8-\frac{(n-1)n(n+1)(n+2)}8 = n(n+1)(n+2)\frac{n+3-(n-1)}8 = \frac{n(n+1)(n+2)}2. $$

In order to find $\sum^n\frac1{T_r}$, notice that: $$ \frac{1}{T_n} = \frac{2}{n(n+1)(n+2)} = \frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)} = A_n - A_{n+1}, $$

where $A_n=\frac{1}{n(n+1)}$. Thus, $$ \sum^n\frac1{T_r} = A_1-A_2 + A_2-A_3 +\ldots +A_{n}-A_{n+1} = A_1-A_{n+1} = \frac12-\frac{1}{(n+1)(n+2)} $$

Answer 2

Hint As you've pointed out, we can recover the original sequence $(T_r)$ from $S_n := \sum_{r = 1}^n T_r$ by $$T_r = S_r - S_{r - 1} = \frac{1}{2} r (r + 1) (r + 2) .$$ Then, $$\sum_{r = 1}^n \frac{1}{T_r} = \sum_{r = 1}^n \frac{2}{r (r + 1) (r + 2)},$$ but we can rewrite this sum in closed form by applying the method of partial fractions.

The partial fraction decomposition of the summand is $$\frac{2}{r (r + 1) (r + 2)} = \frac{1}{r} - \frac{2}{r + 1} + \frac{1}{r + 2} = \left(\frac{1}{r} - \frac{1}{r + 1}\right) - \left(\frac{1}{r + 1} - \frac{1}{r + 2}\right).$$ Thus, the sum telescopes, leaving $$\sum_{r = 1}^n \frac{1}{T_r} = \sum_{r = 1}^n \left[\left(\frac{1}{r} - \frac{1}{r + 1}\right) - \left(\frac{1}{r + 1} - \frac{1}{r + 2}\right)\right] = \frac{1}{2} - \frac{1}{n + 1} + \frac{1}{n + 2} = \color{#df0000}{\boxed{\frac{n (n + 3)}{2 (n + 1) (n + 2)}}} .$$