I'm studying for a Discrete Math exam and I'm preparing myself with an (unsolved) exam from a previous year.

One of those exams had two exercises in a section of the Exam.

The first exercise - that I managed to solve - asked us to calculate the following summation using the Telescoping series method:

$$\sum_{i=0}^n \frac{1}{(i+1)(i+2)}$$ ...which I solved, using "Partial Fraction Decomposition" and the "Telescoping Series" method, reaching the correct result: $$\frac{n+1}{n+2}$$ (I already checked that my answer is correct in "Wolfram Alpha" for this first exercise).

The second exercise, had a similar summation:

$$\sum_{i=1}^n \frac{1}{i(i+1)(i+2)}$$

It only asks to calculate the sum, without specifying a method to use in the calculation. Because it's a rather similar summation from the one in previous exercise, I'm guessing it's possible to use the answer from the previous question as part of the solution. As you can see, the differences in this 2nd exercise are the following:

  • the index of the summation starts with $1$ instead of $0$
  • there's an extra $i$ in the denominator

It makes sense, to me, that this second summation starts with 1 instead of 0, to avoid have the first element of the sum divide by $0$.

Wolfram Alpha "says" the result of this second summation is: $$\frac{n(n+3)}{4(n+1)(n+2)}$$

I'm guessing there should be a way to use the result of the first sum as part of the solution of this second sum, but several experiments that I did to "transform" part of the second sum into the first sum have failed.

Can anyone help or give me some hints, please?


2 Answers 2


\begin{align} {1 \over i\left(i + 1\right)\left(i + 2\right)} &= {1/2 \over i} - {1 \over i + 1} + {1/2 \over i + 2} = {1 \over 2}\left[% \left({1 \over i} - {1 \over i + 1}\right) - \left({1 \over i + 1} - {1 \over i + 2}\right) \right] \\[3mm]&= {1 \over 2}\left[% {1 \over i\left(i + 1\right)} - {1 \over \left(i + 1\right)\left(i + 2\right)} \right] \\[1cm]& \end{align}

\begin{align} \sum_{i = 1}^{n}{1 \over i\left(i + 1\right)\left(i + 2\right)} &= {1 \over 2}\sum_{i = 1}^{n}{1 \over i\left(i + 1\right)} - {1 \over 2}\sum_{i = 1}^{n}{1 \over \left(i + 1\right)\left(i + 2\right)} \\[3mm]&= {1 \over 2}\sum_{i = 0}^{n - 1}{1 \over \left(i + 1\right)\left(i + 2\right)} - \left[% {1 \over 2}\sum_{i = 0}^{n}{1 \over \left(i + 1\right)\left(i + 2\right)} - {1 \over 4} \right] \\[3mm]&= {1 \over 2}\left[% \sum_{i = 0}^{n}{1 \over \left(i + 1\right)\left(i + 2\right)} - {1 \over \left(n + 1\right)\left(n + 2\right)} \right] - {1 \over 2}\left[% \sum_{i = 0}^{n}{1 \over \left(i + 1\right)\left(i + 2\right)} - {1 \over 2} \right] \\[3mm]&= {1 \over 4} - {1 \over 2}\,{1 \over \left(n + 1\right)\left(n + 2\right)} = {\left(n + 1\right)\left(n + 2\right) - 2 \over 4\left(n + 1\right)\left(n + 2\right)} = {n^{2} + 3n \over 4\left(n + 1\right)\left(n + 2\right)} \\[3mm]&= \color{#ff0000}{\large% {n\left(n + 3\right) \over 4\left(n + 1\right)\left(n + 2\right)}} \end{align}


Here is a solution:

Let's denote $$J(0,n)=\sum\limits_0^n\frac{1}{(i+1)(i+2)}=\frac{n+1}{n+1}$$ (you have already established that)

Let's take a look at $$\frac{1}{i(i+1)(i+2)}=\frac{1/2}{i}+\frac{-1}{i+1}+\frac{1/2}{i+2}$$ (this can beobtianed using partial fraction decomppsition for example) Furthermore, we can manipulate with these fractions to get $$\frac{1/2}{i}+\frac{-1}{i+1}+\frac{1/2}{i+2}=\left[\frac{1/2}{i}+\frac{-1/2}{i+1}\right]+\left[\frac{-1/2}{i+1}+\frac{1/2}{i+2}\right]=$$ $$=\frac{1}{2}\frac{1}{i(i+1)}-\frac{1}{2}\frac{1}{(i+1)(i+2)}$$

We want $$\sum_1^n\frac{1}{i(i+1)(i+2)}=\frac{1}{2}\sum_1^n\frac{1}{i(i+1)}-\frac{1}{2}\sum_1^n\frac{1}{(i+1)(i+2)}=$$ $$=\frac{1}{2}\sum_{j=0}^{n-1}\frac{1}{(j+1)(j+2)}-\frac{1}{2}\sum_1^n\frac{1}{(i+1)(i+2)}=\frac{1}{2}J(0,n-1)-\frac{1}{2}J(1,n)=\frac{1}{2}\left[J(0,n-1)-J(1,n)\right]$$ Notice how the middle terms of the difference collapse and we are left with the first term minus the last term of the original sum $J(0,n)$ i.e. $$\frac{1}{2}-\frac{1}{(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}.$$


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