The properties of sum

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

Linearity

$$ \forall \bigl(a_n, b_n\bigr), \enspace \forall (\lambda, \mu) \in \hspace{0.04em} \mathbb{R}^2,$$

$$ \sum_{i \ \in \ I} \Bigl[ \lambda \ a_i + \mu \ b_i \Bigr] = \lambda \sum_{i \ \in \ I} a_i + \mu \sum_{i \ \in \ I} b_i $$

Change of index

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

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
  
  $$ \forall (p, n) \in \hspace{0.04em} \mathbb{N}^2, \enspace p \leqslant n, $$
  
  $$ \sum_{k = p}^n a_k = \sum_{i = 0}^{n - p} a_{p + i} $$
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 :
  
  $$ \forall n \in \hspace{0.04em} \mathbb{N},$$
  
  $$ \sum_{1 \leqslant i \leqslant j \leqslant n}^n \Bigl[ a_{i,j} \Bigr] = \sum_{i = 1}^n \sum_{j = i}^n a_{i, j} = \sum_{j = 1}^n \sum_{i = 1}^j a_{i, j} $$

Telescoping of terms of a recurring numerical series

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

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

Proofs

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:

$$ \sum_{i \ \in \ I} \Bigl[ \lambda \ a_i + \mu \ b_i \Bigr] $$

This sum can be separated into two elements:

$$ \sum_{i \ \in \ I} \Bigl[ \lambda \ a_i + \mu \ b_i \Bigr] = \sum_{i \ \in \ I} \Bigl[ \lambda \ a_i \Bigr] + \sum_{i \ \in \ I} \Bigl[ \mu \ b_i \Bigr] $$

By expanding the sums, we can factor the coefficients:

$$ \sum_{i \ \in \ I} \Bigl[ \lambda \ a_i + \mu \ b_i \Bigr] = \Bigl[ \lambda \ a_{i_1} + \lambda \ a_{i_2} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \lambda \ a_{i_n} \Bigr] + \Bigl[ \mu \ b_{i_1} + \mu \ b_{i_2} + \ ... \ + \hspace{0.2em} \mu \ b_{i_n} \Bigr] $$

$$ \sum_{i \ \in \ I} \Bigl[ \lambda \ a_i + \mu \ b_i \Bigr] = \lambda \ \Bigl[ a_{i_1} + a_{i_2} + \ ... \ + \hspace{0.2em} a_{i_n} \Bigr] + \mu \ \Bigl[ b_{i_1} + b_{i_2} + \ ... \ + \hspace{0.2em} b_{i_n} \Bigr] $$

And finally,

$$ \forall \Bigl[ (a_n)_{n \in \mathbb{N}}, \ (b_n)_{n \in \mathbb{N}} \Bigr], \enspace \forall (\lambda, \mu) \in \hspace{0.04em} \mathbb{R}^2,$$

$$ \sum_{i \ \in \ I} \Bigl[ \lambda \ a_i + \mu \ b_i \Bigr] = \lambda \sum_{i \ \in \ I} a_i + \mu \sum_{i \ \in \ I} b_i $$

Change of index

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:
  
  $$ \sum_{k = 0}^n a_k = a_0 + a_1 + \ ... \ + \hspace{0.2em} a_{n - 1} + a_n $$
  
  Then we reverse the direction:
  
  $$ \sum_{k = 0}^n a_k = a_n + a_{n - 1} + \ ... \ + \hspace{0.2em} a_1 + a_0 $$
  
  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:
  
  $$ \sum_{k = p}^n a_k = a_p + a_{p + 1} + \ ... \ + \hspace{0.2em} a_{n - 1} + a_n $$
  
  Arranging the indexes, we have:
  
  $$ \sum_{k = p}^n a_k = a_{p + 0} + a_{p + 1} + \ ... \ + \hspace{0.2em} a_{p + (n - p - 1)} + a_{p + (n - p)} $$
  
  And finally,
  
  $$ \forall (p, n) \in \hspace{0.04em} \mathbb{N}^2, \enspace p \leqslant n, $$
  
  $$ \sum_{k = p}^n a_k = \sum_{i = 0}^{n - p} a_{p + i} $$

Double sum

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

Sum which not depend of indexes

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

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

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

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

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 $j$	$$ 1 $$	$$ 2 $$	$$ 3 $$	$$ ... $$	$$ n $$
$$ 1 $$	$$ a_{1,1} $$	$$ a_{1,2} $$	$$ a_{1,3} $$	$$ ... $$	$$ a_{1,n} $$
$$ 2 $$	$$ a_{2,1} $$	$$ a_{2,2} $$	$$ a_{2,3} $$	$$ ... $$	$$ a_{2,n} $$
$$ 3 $$	$$ a_{3,1} $$	$$ a_{3,2} $$	$$ a_{3,3} $$	$$ ... $$	$$ a_{3,n} $$
$$ ... $$	$$ ... $$	$$ ... $$	$$ ... $$	$$ ... $$	$$ ... $$
$$ m $$	$$ a_{m,1} $$	$$ a_{m,2} $$	$$ a_{m,3} $$	$$ ... $$	$$ a_{m,n} $$

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) $.

Line vision

By adopting a line vision, we have:

$$ \sum_{1 \leqslant i \leqslant j \leqslant n}^n \Bigl[ a_{i,j} \Bigr] = \textcolor{rgb(118 139 240)}{a_{1,1} + a_{1,2} + a_{1,3} + \ ... \ + \hspace{0.2em} a_{1,n}} + \textcolor{rgb(93 183 129)}{a_{2,2} + a_{2,3} \hspace{0.2em} + \ ... \ + \hspace{0.2em} a_{2,n}} + \textcolor{rgb(232 124 124)}{a_{3,3} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_{3,n}} + \textcolor{rgb(197 144 218)}{a_{n,n}} $$

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 $(j = i)$ :

$$ \sum_{1 \leqslant i \leqslant j \leqslant n}^n \Bigl[ a_{i,j} \Bigr] = \textcolor{rgb(118 139 240)}{\sum_{j = 1}^n a_{1,j}} + \textcolor{rgb(93 183 129)}{\sum_{j = 2}^n a_{2,j}} + \textcolor{rgb(232 124 124)}{\sum_{j = 3}^n a_{3,j}} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \textcolor{rgb(213 101 255)}{\sum_{j = n}^n a_{n,j}} $$

And as a result,

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

Column vision

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

$$ \sum_{1 \leqslant i \leqslant j \leqslant n}^n \Bigl[ a_{i,j} \Bigr] = \textcolor{rgb(118 139 240)}{a_{1,1}} + \textcolor{rgb(54 152 46)}{a_{1,2} + a_{2,2}} + \textcolor{rgb(232 124 124)}{a_{1,3} + a_{2,3} + a_{3,3}} + \ ... \ + \hspace{0.2em} \textcolor{rgb(197 144 218)}{a_{1,n} + a_{2,n} + a_{3,n} + \ ... \ + \hspace{0.2em} a_{n,n}} $$

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

$$ \sum_{1 \leqslant i \leqslant j \leqslant n}^n \Bigl[ a_{i,j} \Bigr] = \textcolor{rgb(118 139 240)}{\sum_{i = 1}^1 a_{i,1}} + \textcolor{rgb(93 183 129)}{\sum_{i = 1}^2 a_{i,2}} + \textcolor{rgb(232 124 124)}{\sum_{i = 1}^3 a_{i,3}} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \textcolor{rgb(213 101 255)}{\sum_{i = 1}^n a_{i,n}} $$

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 s depends 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,

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

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

$$ \sum_{k=n_0}^n \Bigl [a_{k+1} - a_k \Bigr] = a_{k+1} + \underbrace{a_{k} - a_{k}} _\text{ $= 0$} + \underbrace{ a_{k-1} - a_{k-1}} _\text{ $= 0$} + \ ... \ + \underbrace{a_2 - a_2} _\text{ $= 0$} + \underbrace{a_1 - a_1} _\text{ $= 0$} - a_0 $$

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

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

Example

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

Proofs

Linearity

Change of index

Simple sum

Reversing direction

Shift to zero

Double sum

Sum which not depend of indexes

Sum which do depend of indexes

Line vision

Column vision

Telescoping of terms of a recurring numerical series

Example

Telescoping of terms

Partial fraction decomposition

Calculation of the partial sum with telescoping