I am trying to find the partial sum formula of the following series:

$$ \sum_{y=1}^{\infty} \frac{4y^2-12y+9}{(y+3)(y+2)(y+1)y} $$

I have tried using Faulhaber's formula without success. I have also tried rewriting the system using partial fraction decomposition to obtain 4 terms. This didn't solve the issue either.

When I use WolframAlpha to evaluate the sum, (or other computational software), it becomes 1/2. Is there some way to derive the this infinite sum to a partial sum formula?

This partial sum formula is according to WolframAlpha: $$ \frac{n^3-2n^2+3n}{2(n+1)(n+2)(n+3)} $$

Thank you in advance! J


Decompose the fraction on simple elements:

$$\frac{4y^2-12y+9}{(y+3)(y+2)(y+1)y}=\frac{a}{y}+\frac{b}{y+1}+\frac{c}{y+2}+\frac{d}{y+3}$$ and since the series is convergent then we have $a+b+c+d=0$. Now the partial sum is

$$\sum_{y=1}^n \frac{a}{y}+\frac{b}{y+1}+\frac{c}{y+2}+\frac{d}{y+3}=a\sum_{y=1}^n \frac{1}{y}+b\sum_{y=2}^{n+2} \frac{1}{y}+c\sum_{y=3}^{n+3} \frac{1}{y}+d\sum_{y=4}^{n+4} \frac{1}{y}$$ and the simplification is clear. Can you take it from here?


Yes I managed to simplify the question as you have. I got that

$$ -13.5\sum_{y=1}^{\infty} \frac{1}{y+3} + 24.5\sum_{y=1}^{\infty} \frac{1}{y+2} -12.5\sum_{y=1}^{\infty} \frac{1}{y+1} + 1.5\sum_{y=1}^{\infty} \frac{1}{y} $$

However, I can't figure out how to find the partial sum formula of each term (every summation term). The series seem harmonic and as I understand it is rather difficult to find an explicit formula for its partial sum?

Moreover, I can see that you have introduced n+1, n+2, ... and so on on the summation series. How was that done? Why haven't you just used n? And then let $$n\rightarrow\infty$$


  • $\begingroup$ You're not supposed to figure out each individually, you're supposed to see how the first and third series taken together kill almost all terms of the second and fourth one, save a few. $\endgroup$ – Lucian Oct 1 '14 at 15:19
  • $\begingroup$ yea, but how does the series evaluate to 0.5 as n->infinity ? $\endgroup$ – Jam Oct 1 '14 at 15:30
  • $\begingroup$ OK, write the first $10$ terms of each sum explicitly, and then notice how, starting with denominator $=4$, all terms cancel each other out. $\endgroup$ – Lucian Oct 1 '14 at 15:36
  • $\begingroup$ Ok, I now I got it. Thanks so much! $\endgroup$ – Jam Oct 1 '14 at 15:58

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