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The properties of sum

Let \( \bigl(a_n, b_n\bigr) \) be two numerical sequences.

Let \( n \in \hspace{0.04em} \mathbb{N} \) be a natural number.

1. Simple sum

1. Reversing direction

$$ \forall n \in \hspace{0.04em} \mathbb{N},$$
$$ \sum_{k = 0}^n a_k = \sum_{i = 0}^n a_{n - i} $$

2. Shift to zero

$$ \forall (p, n) \in \hspace{0.04em} \mathbb{N}^2, \enspace p \leqslant n, $$
$$ \sum_{k = p}^n a_k = \sum_{i = 0}^{n - p} a_{p + i} $$

2. Double sum

1. Sum which not depend of indexes

In the case of a double sum indexed by a rectangle \( [\![1,m]\!] \times [\![1,n]\!] \):

$$ \forall (m,n) \in \hspace{0.04em} \mathbb{N}^2,$$
$$ \sum_{i = 1}^m \sum_{j = 1}^n a_{i,j} = \sum_{j = 1}^n \sum_{i = 1}^m a_{i, j} $$

2. Sum which do depend of indexes

In the case of a double triangular sum :

$$ \forall n \in \hspace{0.04em} \mathbb{N},$$
$$ \sum_{1 \leqslant i \leqslant j \leqslant n}^n \Bigl[ a_{i,j} \Bigr] = \sum_{i = 1}^n \sum_{j = i}^n a_{i, j} = \sum_{j = 1}^n \sum_{i = 1}^j a_{i, j} $$

When subtracting two consecutive terms within a sum, we can perform a telescoping of terms:

Demonstrations

Let \( \bigl(a_n, b_n\bigr) \) be two numerical sequences.

Linearity

Let be \( (\lambda , \mu) \in \hspace{0.04em} \mathbb{R}^2 \) two real numbers. We start with the following sum:

This sum can be separated into two elements:

By expanding the sums, we can factor the coefficients:

And finally,

Change of index

1. Simple sum

Let \( n \in \hspace{0.04em} \mathbb{N} \) be a natural number.

1. Reversing direction

Let \( p \in \hspace{0.04em} \mathbb{N} \) be a natural number with \((p \leqslant n)\).

We start with the following sum:

$$ \sum_{k = 0}^n a_k = a_0 + a_1 \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_{n - 1} + a_n $$

Then we reverse the direction:

$$ \sum_{k = 0}^n a_k = a_n + a_{n - 1} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_1 + a_0 $$

And finally,

$$ \forall n \in \hspace{0.04em} \mathbb{N},$$
$$ \sum_{k = 0}^n a_k = \sum_{i = 0}^n a_{n - i} $$

2. Shift to zero

We start with the following sum:

$$ \sum_{k = p}^n a_k = a_p + a_{p + 1} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_{n - 1} + a_n $$

Arranging the indexes, we have:

$$ \sum_{k = p}^n a_k = a_{p + 0} + a_{p + 1} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_{p + (n - p - 1)} + a_{p + (n - p)} $$

And finally,

$$ \forall (p, n) \in \hspace{0.04em} \mathbb{N}^2, \enspace p \leqslant n, $$
$$ \sum_{k = p}^n a_k = \sum_{i = 0}^{n - p} a_{p + i} $$

2. Double sum

Let \( (m,n) \in \hspace{0.04em} \mathbb{N}^2 \) be two natural numbers.

1. Sum which not depend of indexes

We start from the following double sum, indexed by a rectangle \( [\![1,m]\!] \times [\![1,n]\!] \):

$$ \sum_{i = 1}^m \sum_{j = 1}^n a_{i,j} $$
$$ \sum_{i = 1}^m \sum_{j = 1}^n a_{i,j} = \sum_{i = 1}^m \left( a_{i, 1} + a_{i, 2} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_{i, n} \right) $$
$$ \sum_{i = 1}^m \sum_{j = 1}^n a_{i,j} = \sum_{i = 1}^m a_{i, 1} + \sum_{i = 1}^m a_{i, 2} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \sum_{i = 1}^m a_{i, m} $$

We notice that it is the same sum by reversing the indices.

So, for double sums where each of the sums does not depend on the indexes, we can easily switch the sum symbols:

$$ \sum_{i = 1}^m \sum_{j = 1}^n a_{i,j} = \sum_{j = 1}^n \sum_{i = 1}^m a_{i, j} $$

De manière visuelle, on peu voir que faire la sommes des lignes ou la somme des colonnes reviendra au même.

Indice \(i\) Indice \(j\)	$$ 1 $$	$$ 2 $$	$$ 3 $$	$$ ... $$	$$ n $$
$$ 1 $$	$$ a_{1,1} $$	$$ a_{1,2} $$	$$ a_{1,3} $$	$$ ... $$	$$ a_{1,n} $$
$$ 2 $$	$$ a_{2,1} $$	$$ a_{2,2} $$	$$ a_{2,3} $$	$$ ... $$	$$ a_{2,n} $$
$$ 3 $$	$$ a_{3,1} $$	$$ a_{3,2} $$	$$ a_{3,3} $$	$$ ... $$	$$ a_{3,n} $$
$$ ... $$	$$ ... $$	$$ ... $$	$$ ... $$	$$ ... $$	$$ ... $$
$$ m $$	$$ a_{m,1} $$	$$ a_{m,2} $$	$$ a_{m,3} $$	$$ ... $$	$$ a_{m,n} $$

2. Sum which do depend of indexes

We start from the following double triangular sum:

$$ \sum_{1 \leqslant i \leqslant j \leqslant n}^n \Bigl[ a_{i,j} \Bigr] $$

It is the double sum where we always have \( (i \leqslant j) \).

1. Line vision

By adopting a line vision, we have:

$$ \sum_{1 \leqslant i \leqslant j \leqslant n}^n \Bigl[ a_{i,j} \Bigr] = \textcolor{#606B9E}{a_{1,1} + a_{1,2} + a_{1,3} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_{1,n}} + \textcolor{#4A8051}{a_{2,2} + a_{2,3} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_{2,n}} + \textcolor{#AF5F5F}{a_{3,3} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_{3,n}} + \textcolor{#7C578A}{a_{n,n}} $$

Indice \(i\) Indice \(j\)	$$ 1 $$	$$ 2 $$	$$ 3 $$	$$ ... $$	$$ n $$
$$ 1 $$	$$ a_{1,1} $$	$$ a_{1,2} $$	$$ a_{1,3} $$	$$ ... $$	$$ a_{1,n} $$
$$ 2 $$	$$ $$	$$ a_{2,2} $$	$$ a_{2,3} $$	$$ ... $$	$$ a_{2,n} $$
$$ 3 $$	$$ $$	$$ $$	$$ a_{3,3} $$	$$ ... $$	$$ a_{3,n} $$
$$ ... $$	$$ $$	$$ $$	$$ $$	$$ ... $$	$$ ... $$
$$ n $$	$$ $$	$$ $$	$$ $$	$$ $$	$$ a_{n,n} $$

All these elements are the sum closed on the index \(i\), in which each sum-term starts at \((j = i)\) :

$$ \sum_{1 \leqslant i \leqslant j \leqslant n}^n \Bigl[ a_{i,j} \Bigr] = \textcolor{#606B9E}{\sum_{j = 1}^n a_{1,j}} + \textcolor{#4A8051}{\sum_{j = 2}^n a_{2,j}} + \textcolor{#AF5F5F}{\sum_{j = 3}^n a_{3,j}} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \textcolor{#7C578A}{\sum_{j = n}^n a_{n,j}} $$

And as a result,

$$ \sum_{1 \leqslant i \leqslant j \leqslant n}^n \Bigl[ a_{i,j} \Bigr] = \sum_{i = 1}^n \sum_{j = i}^n a_{i, j} $$

2. Column vision

By adopting a column view, we obtain the sum in this form:

$$ \sum_{1 \leqslant i \leqslant j \leqslant n}^n \Bigl[ a_{i,j} \Bigr] = \textcolor{#606B9E}{a_{1,1}} + \textcolor{#4A8051}{a_{1,2} + a_{2,2}} + \textcolor{#AF5F5F}{a_{1,3} + a_{2,3} + a_{3,3}} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \textcolor{#7C578A}{a_{1,n} + a_{2,n} + a_{3,n} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_{n,n}} $$

All these elements are the sum closed on the index \(j\), in which each sum term goes at most to \((i = j)\) :

$$ \sum_{1 \leqslant i \leqslant j \leqslant n}^n \Bigl[ a_{i,j} \Bigr] = \textcolor{#606B9E}{\sum_{i = 1}^1 a_{i,1}} + \textcolor{#4A8051}{\sum_{i = 1}^2 a_{i,2}} + \textcolor{#AF5F5F}{\sum_{i = 1}^3 a_{i,3}} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \textcolor{#7C578A}{\sum_{i = 1}^n a_{i,n}} $$

And as a result,

$$ \sum_{1 \leqslant i \leqslant j \leqslant n}^n \Bigl[ a_{i,j} \Bigr] = \sum_{j = 1}^n \sum_{i = 1}^j a_{i, j} $$

So, for double sums where each of the sum sdepends on indexes, we can make the following change of variable:

$$ \forall n \in \hspace{0.04em} \mathbb{N},$$
$$ \sum_{1 \leqslant i \leqslant j \leqslant n}^n \Bigl[ a_{i,j} \Bigr] = \sum_{i = 1}^n \sum_{j = i}^n a_{i, j} = \sum_{j = 1}^n \sum_{i = 1}^j a_{i, j} $$

Telescoping of terms of a recurring numerical series

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

We will have,

Arranging this expression, the terms will be annihilated one by one.

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

Example

1. Telescoping of terms

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:

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

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.

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

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