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For $r\in \mathbb N$, find $f(r)$ such that $$f(r)-f(r-1)=\frac {r+2}{r(r+1)(r+3)}$$

i found this on one of my class paper's. it was under the progression addition exercises. i tried many different approach's. but it seems nothing works. is there a easy way to calculate this. I tried to multiply both numerator and denominator by (r+2) and taking the numerator as Ar^2+Br+C with three factors in the denominator and subtracting them such that f(r)-f(r-1) but it didn't work and it seems to be really complicated and long. And i also tries to find the partial fractions but this has 3 partial factors so i couldn't proceed with that approach either.

lulu
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Chamitha Senaratne
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    Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community May 21 '25 at 17:13
  • This is hard to follow. Do you mean $f(r)-f(r-1)=\frac {r+2}{r(r+1)(r+3)}$? Something else? In any case, since $f$ is only determined up to a constant, you might as well take $f(0)=0$. In that case, you might at least be able to get $f(r)$ for integer $r$. Not sure if $r$ was meant to be an integer or a real number, or something else entirely. – lulu May 21 '25 at 17:17
  • yes that is correct. (r is any positive integer) – Chamitha Senaratne May 21 '25 at 17:29
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    Similar problems : https://math.stackexchange.com/q/2323294/1157207, https://math.stackexchange.com/q/3057821/1157207, https://math.stackexchange.com/q/3967216/1157207, – Amrut Ayan May 21 '25 at 17:29
  • Please edit to include that crucial information in your post. Also, given my comment, this amounts to simply summing this expression. It's a bit tedious, but not difficult. – lulu May 21 '25 at 17:30
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    Hint: telescoping summation. – Abezhiko May 21 '25 at 17:51

1 Answers1

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We have the recurrence:

$$ f(r) - f(r-1) = \frac{r + 2}{r(r + 1)(r + 3)} $$

The first step is to decompose this into fractions such that

$$ \frac{r+2}{r(r+1)(r+3)} = \frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+3} $$

After multiplying the denominators and solving the linear system, we obtain $A = 2/3$, $B = -1/2$ and $C = -1/6$.

All in all,

$$ \frac{r+2}{r(r+1)(r+3)} = \frac{2}{3r} - \frac{1}{2(r+1)} - \frac{1}{6(r+3)}. $$

Next we exploit the telescoping of the relation, note that

$$\sum_{k=1}^r f(k) - f(k-1) = f(r) - f(0)$$

Let us take $f(0) = 0$ for the sake of simplicity. Replacing $f(k) - f(k-1)$ with our above decomposition yields

$$f(r) = \sum_{k=1}^r \left(\frac{2}{3k} - \frac{1}{2(k+1)} - \frac{1}{6(k+3)}\right)$$

Now, write out the first few terms of the sum, and the last few (the latter in terms of $r$) to demonstrate telescoping. In the end you should obtain:

$$f(r) = \frac{29}{36} - \frac{1}{6(r+2)} - \frac{1}{6(r+3)} - \frac{2}{3(r+1)}$$

mizuoto
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