Calculate $\sum\limits_{i=0}^{n-1}\frac{1}{(i+1)(i+2)}$ in use of disturbance of sum 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:
$$ S_n=\sum_{k=0}^{n}q^k = ? $$
$$ S_n+q^{n+1}=1+\sum_{k=1}^{n+1}q^k=1+\sum_{k=0}^{n}q^{k+1}= 1+q\sum_{k=0}^{n}q^{k}=1+qS_n
$$
$$(q-1)S_n=q^{n+1}-1
$$
$$ S_n=\frac{q^{n+1}-1}{q-1}
$$
 A: Telescoping is a method where we try to write $a_{k}$ on the form $a_k = b_{k+1} - b_k$ for some sequence $b_k$. From this form the sum of the series follows directly as $\sum_{k=0}^n a_k = b_{n+1} - b_0$.
What you are trying to do, from looking at your example, looks like the opposite of telescoping: to try to write $a_k$ on a form like $a_{k+1} - qa_k = b_k$ where $q$ is some constant (this can be generalized, but let's stick with the simplest case here). If you are able to do so and where $b_k$ is 'easier' than $a_k$ so that you are able to sum $b_k$ then by summing the recurrence relation you would be able to find the sum of $a_k$ as
$$\sum_{k=0}^n a_{k} = \frac{a_0 - a_{n+1} + \sum_{k=0}^n b_k}{1-q}$$
However, don't expect all methods to work with all series. Just as telescoping, this approach would only work in some special cases. Being good at finding sums is to figure out what methods works best with what series. For the series in question telescoping is definitely the way to go and the latter method would not work as $b_k$ would be an even more complicated series than that of $a_k$ so you should just discard this method for the given series.
A: 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 $$\sum_{k=m}^M (a_k - a_{k+1}) = a_m - a_{M+1}.$$
