The telescoping of terms in recurring numerical series

Let \( (a_n)_{n \in \mathbb{N}} \) a numerical sequence.

When we want to calculate the series \( \sum \bigl [a_{k+1} - a_k \bigr] \), there will be a telescoping of terms such as:

$$\sum_{k=0}^n \Bigl [a_{k+1} - a_k \Bigr] = a_{n+1} - a_{0} $$

All that will remain is that the last minus the first of the series.

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Demonstration

We want to calculate the series \( \sum \bigl [a_{k+1} - a_k \bigr] \) from \( k = 0 \) to \( n \).

We will have,

$$ \sum_{k=n_0}^n \Bigl [a_{k+1} - a_k \Bigr] = a_{1} - a_{0} + a_{2} - a_{1} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_{n_{k}} - a_{n_{k-1}} + a_{n_{k+1}} - a_{n_{k}} $$

The terms will be annihilated one by one.

$$ \sum_{k=n_0}^n \Bigl [a_{k+1} - a_k \Bigr] = a_{k+1} + a_{k} - a_{k} + a_{k-1} - a_{k-1} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} + a_2 - a_2 + a_1 - a_1 - a_0 $$

All that will remain is that the last minus the first of the series.

$$\sum_{k=0}^n \Bigl [a_{k+1} - a_k \Bigr] = \underbrace{a_{n+1}} _\text{first term} - \underbrace{a_{0}} _\text{last term} $$

As a result we do have:

$$\sum_{k=0}^n \Bigl [a_{k+1} - a_k \Bigr] = a_{n+1} - a_{0} $$

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Example

Let us calculate the partial sum of the following series:

$$S_n = \sum_{k=1}^n \frac{1}{k(k+1)}$$

To carry out his calculation, we first need to turn this fraction into a partial fraction decomposition.

1. Partial fraction decomposition

Let set the \(F(X) \) function down:

$$F(X) = \frac{1}{X(X+1)} \qquad (F(X))$$

We are looking for two reals \( a \) and \(b\) such as:

$$F(X) = \frac{a}{X} + \frac{b}{X+1}$$

Putting in the same dénominateur, we do have:

$$F(X) = \frac{a(X+1) + b X}{X(X+1)} \qquad (\tilde{F}(X)) $$

The idea here is to use both forms \( (F(X)) \) and \( (\tilde{F}(X)) \) to obtain an equivalence and determine \( a \) and \(b\), we do have:

$$F(X) X = \frac{1}{(X+1)} \qquad (F(X))$$

$$F(X) X = \frac{a(X+1) + b X}{(X+1)} \qquad (\tilde{F}(X)) $$

So,

$$ F(X) X = \frac{1}{(X+1)}= \frac{a(X+1) + b X}{(X+1)} $$

$$ F(X) X = \frac{1}{(X+1)}= a + \frac{ b X}{(X+1)} $$




Doing \( (X = 0)\), we determine \( a \):

$$ \underset{(X=0)}{F(X)} X = \frac{1}{(X+1)}= a \Longrightarrow (a = 1) $$

We can do the same thong to determine \(b\), doing \( (X = -1)\) it will remain \( b \):

$$ \underset{(X=-1)}{F(X)} (X+1) = \frac{1}{X}= b \Longrightarrow (b = - 1) $$

We then have our couple of solutions:

$$ \Biggl \{ \begin{align*} a = 1 \\ b = -1 \end{align*} $$

Thus, \(F(X) \) can be written:

$$F(X) = \frac{1}{X} - \frac{1}{X+1}$$

2. Calculation of the partial sum with telescoping

Thanks to the partial fraction decomposition, we do have now:

$$F(X) = \frac{1}{X(X+1)} =\frac{1}{X} - \frac{1}{X+1}$$

Our series:

$$S_n = \sum_{k=1}^n \frac{1}{k(k+1)}$$

becomes,

$$S_n = \sum_{k=0}^n \Biggl[ \frac{1}{k} - \frac{1}{k+1} \Biggr]$$

We extract the \((-)\) sign to have a sequence under the form \( \bigl [a_{k+1} - a_k \bigr] \).

$$S_n = -\sum_{k=1}^n \Biggl[ \frac{1}{k+1} -\frac{1}{k} \Biggr]$$

Let set down:

$$ a_k = \frac{1}{k} $$

to obtain,

$$\sum_{k=1}^n \Biggl[ \frac{1}{k+1} -\frac{1}{k} \Biggr] = \sum_{k=0}^n \Bigl [a_{k+1} - a_k \Bigr]$$




Then we apply:

$$\sum_{k=0}^n \Bigl [a_{k+1} - a_k \Bigr] = a_{n+1} - a_{0} $$

We can now carry out the telescoping.

$$S_n = -\Biggl[ \frac{1}{n+1} -\frac{1}{1} \Biggr]$$

$$S_n = 1 -\frac{1}{n+1} $$