Algebraic identities in series

Sum of an arithmetic series… within the reach of a Gauss ========================================================================================================

An arithmetic sequence is defined by its initial term, a₀, and by its common difference, r. Each term of the sequence is equal to the previous term plus the common difference. Thus, the first term is equal to a₁ = a₀ + r, the next term to a₂ = a₀ + r + r = a₀ + 2r, and so on. The nth term is equal to *a_n = a*₀ + n × r.

How can we calculate a partial sum (the sum of the first terms) of an arithmetic series (a series whose terms are in arithmetic progression)? A solution of striking simplicity is credited, so mathematical legend has it, to Carl Friedrich Gauss, back when he was just a child. Finding him too restless in class, a teacher, to get some peace, asked him to calculate the sum of the first hundred integers. Imagine the teacher's surprise on getting the answer just a few seconds later! Gauss had written the integers on the first line in increasing order, and on the second line in decreasing order:

1 2 3 4 ... 99 100

100 99 98 97 ... 2 1

By summing the hundred columns (each totaling 101), he very quickly arrived at the conclusion that the sum in question was equal to half of 10,100, that is, 5,050.

Generalizing this method makes it possible to show that the sum S*_n* of the terms of the previous series (a₀ + a₁ + a₂ + a₃ + a₄ +… *a_n*) is equal to

**Sum of a geometric series**

A geometric sequence is likewise defined by its initial term a₀ and by its common ratio, r. But this time, each term of the sequence is equal to the previous term multiplied by the common ratio. Thus, the nth term is equal to *a_n = a*₀ *rⁿ*.

How can we calculate its partial sum S*_n = a*₀ + a₁ + a₂ + a₃ + a₄ + … *a_n ? There are two cases. The first is trivial: it is the case where the common ratio is 1. The sequence is constant, and the sum of the terms from 0 to n is equal to (n + 1) × a*₀. In the general case, the trick is to multiply S*_n by (1 – r*). We then get:

(1 – r) S*_n = a*₀ (1 – r)(1 + r + r² + r³ + r⁴ + … + *rⁿ*).

Expanding the product, we find: (1 – r)(1 + r + r² + r³ + r⁴ + … + *rⁿ), that the terms cancel out in pairs, leaving only 1 – rⁿ*⁺¹. Thus,

Can we calculate the sum of an infinite number of real numbers? The answer is… "sometimes." Starting with the simplest cases, the two progressions (arithmetic and geometric):

• the sum of a nonzero arithmetic series always has an infinite limit (+∞ or –∞);

• the sum of a nonzero geometric series has a finite limit if and only if the absolute value of its common ratio is strictly less than 1; in that case, *r ⁿ*⁺¹ tends to 0 as n grows, and the sum will get closer and closer to the limit

The question can be broadened to other series. We then need to draw on real analysis to reach the conclusion that a sum S exists (the limit of the partial sums). This is called the convergence of the series.

Careful: it is sometimes possible to show that this limit exists without necessarily knowing its value! Certain results tied to the topological structure of the real numbers will often be needed to reach this conclusion. Here is the simplest of them: An increasing, bounded-above sequence of real numbers always has a limit.

Thus, a series of positive numbers whose partial sums are bounded above is always convergent. Consequence: if the terms *u_n of a series are bounded above by the terms v_n* of a convergent series, then the series $\sum u_n\backslash sum\; u\_n$ is convergent.

We can therefore use the convergence of any geometric series with positive common ratio r < 1 to reach a very useful result: if the ratio of two consecutive terms of a series with positive terms $\frac{u_{n+1}}{u_n}\backslash frac\{u\_\{n+1\}\}\{u\_n\}$ stays below a number r strictly less than 1, then the series $\sum u_n\backslash sum\; u\_n$ is convergent.