TITLE: Calculate $\sum\limits_{i=0}^{n-1}\frac{1}{(i+1)(i+2)}$ in use of disturbance of sum
QUESTION [6 upvotes]: I need calculate
$$\sum_{i=0}^{n-1}\frac{1}{(i+1)(i+2)}$$
in use of disturbance of sum method:
$$\boxed{\boxed{\sum_{k=m}^Ma_k+a_{M+1}=a_m+\sum_{k=m+1}^Ma_{k+1}}}
 $$
My try
I know how to solve this task in "traditional" way but I am trying to do it with that method:
$$S_n=S_n$$
$$ \sum_{i=0}^{n-1}\frac{1}{(i+1)(i+2)} + \frac{1}{(n+1)(n+2)} = \frac{1}{2} + \sum_{i=1}^{n}\frac{1}{(i+1)(i+2)}  $$
ok - now I want to have the same piece as on the left side of equal. So i decide to use:
$$  i' = i+1 $$
$$ \frac{1}{2} + \sum_{i=1}^{n}\frac{1}{(i+1)(i+2)} = \frac{1}{2} + \sum_{i=0}^{n-1}\frac{1}{i(i+1)} $$
but there I divide by 0... How to avoid that problem?
Method
I don't know if the name "disturbance of sum" is correct in english so I am going to give example of another task with this approach:
$$ S_n=\sum_{k=0}^{n}q^k = ? $$
$$ S_n+q^{n+1}=1+\sum_{k=1}^{n+1}q^k=1+\sum_{k=0}^{n}q^{k+1}= 1+q\sum_{k=0}^{n}q^{k}=1+qS_n
$$
$$(q-1)S_n=q^{n+1}-1
$$
$$ S_n=\frac{q^{n+1}-1}{q-1}
$$

REPLY [3 votes]: In fact, the intended use of the provided identity $$\sum_{k=m}^M a_k + a_{M+1} = a_m + \sum_{k=m}^M a_{k+1}$$ is not by assigning $$a_k = \frac{1}{(k+1)(k+2)},$$ but rather, $$\boxed{\boxed{a_k = \frac{1}{k+1}}}.$$  When you do this, you immediately obtain (for $m = 0$ and $M = n-1$) $$\sum_{k=0}^{n-1} \frac{1}{k+1} + \frac{1}{n} = 1 + \sum_{k=0}^{n-1} \frac{1}{k+2}.$$  Upon rearrangement, we get $$\sum_{k=0}^{n-1} \frac{1}{k+1} - \frac{1}{k+2} = 1 - \frac{1}{n},$$ and since $$\frac{1}{k+1} - \frac{1}{k+2} = \frac{(k+2) - (k+1)}{(k+1)(k+2)} = \frac{1}{(k+1)(k+2)},$$ this concludes the proof.
Another way to see this is to rewrite the perturbation identity in the form $$\sum_{k=m}^M (a_k - a_{k+1}) = a_m - a_{M+1}.$$