O que seria uma série?

Read · 3 min

By now, you should be familiar with summation notation: given a finite sequence of numbers a1,…,ana_{1}, \ldots, a_{n}, their sum can be written as ∑k=1nak\sum_{k=1}^{n} a_{k}.

As the basic principle behind calculus is to use a limiting process to study infinities, a natural question is to ask whether there is a way to generalize the process of summing a finite sequence of numbers (which is a trivial thing to do) to summing an infinite sequence of numbers.

More precisely, given an infinite sequence a1,a2,…,ak,…a_{1}, a_{2}, \ldots, a_{k}, \ldots, is there some sort of way to rigorously define and make sense of the sum \\ ∑k=1∞ak\sum_{k=1}^{\infty} a_{k}? We will call such an infinite sum an infinite series. (Often we will just drop the adjective infinite and refer to such a sum as a series.)

If we only need to sum a finite sequence of numbers, we simply sum all the numbers in the sequence and declare that value (trivially) to be the sum. We can try to approximate this the best we can with the sum of an infinite sequence of numbers, where we ask what the sum of the first nn terms of the infinite sequence is, and attempt to understand what the behavior of this partial sum is as n→∞n \rightarrow \infty. Let's look at an example to make this precise.

Suppose an=1/3na_{n} = 1/3^{n}. Then the sum of the first nn terms of this sequence, which we will call sns_{n}, for small values of nn, is equal to

s1=a1=1/3s_{1} = a_{1} = 1/3,

s2=a1+a2=1/3+1/9=4/9s_{2} = a_{1} + a_{2} = 1/3 + 1/9 = 4/9,

s3=a1+a2+a3=1/3+1/9+1/27=13/27s_{3} = a_{1} + a_{2} + a_{3} = 1/3 + 1/9 + 1/27 = 13/27, etc.

It is clear that for any nn, we could compute sns_{n}. Furthermore, as n→∞n \rightarrow \infty, sns_{n} contains more and more terms of the 'infinite sum' ∑n=1∞an\sum_{n=1}^{\infty} a_{n}. So we will define the value of the series ∑n=1∞an\sum_{n=1}^{\infty} a_{n} to be the limit of the sequence sns_{n}, if such a limit exists. If the limit of the sequence of partial sums exist, we say that the series ∑n=1∞an\sum_{n=1}^{\infty} a_{n} converges. If the limit of the sequence of partial sums does not exist, we say that the series ∑n=1∞an\sum_{n=1}^{\infty} a_{n} diverges.

Returning to our example where an=1/3na_{n} = 1/3^{n}, can we determine whether ∑n=1∞1/3n\sum_{n=1}^{\infty} 1/3^{n} converges or diverges? If we had an explicit formula for sns_{n}, then we might be able to more easily answer this question. Fortunately, with a little bit of algebra, we can do just that. Recall the identity

(xn−yn)=(x−y)(xn−1+xn−2y+…+yn−1)(x^{n} - y^{n}) = (x-y)(x^{n-1} + x^{n-2}y + \ldots + y^{n-1}). That rightmost term looks an awful lot like our partial sums. If we let x=1,y=1/3x = 1, y = 1/3, then we get

(1−13n)=(1−13)(1+13+…+13n−1).\left(1 - \frac{1}{3^{n}}\right) = \left(1 - \frac{1}{3}\right) \left(1 + \frac{1}{3} + \ldots + \frac{1}{3^{n-1}}\right).

With this substitution, the rightmost term is just 33 times the partial sum sns_{n}. In other words,

(1−13n)=23⋅3sn=2sn,(1 - \frac{1}{3^{n}}) = \frac{2}{3} \cdot 3 s_{n} = 2 s_{n}, or

sn=12(1−13n).s_{n} = \frac{1}{2} (1 - \frac{1}{3^{n}}).

Clearly, as n→∞n \rightarrow \infty, sn→1/2s_{n} \rightarrow 1/2. So we say that the infinite series ∑n=1∞1/3n\sum_{n=1}^{\infty} 1/3^{n} converges, and that its value is 1/21/2.

Notice that the above technique works more generally on any infinite series of the form ∑n=0∞arn=a+ar+ar2+ar3+…\sum_{n=0}^{\infty} ar^{n} = a + ar + ar^{2} + ar^{3} + \ldots. (Notice that the starting index of summation is now 00 instead of 11.) Any series of this form is called a geometric series, and our example was one such series, with a=r=1/3a = r = 1/3. The quantity aa is sometimes called the initial term of the series, and the quantity rr the ratio.

Exercise 1. Show that for r≠1r \neq 1,

a+ar+ar2+…+arn−1=a(1−rn)1−r.a + ar + ar^{2} + \ldots + a r^{n-1} = \frac{a(1 - r^{n})}{1 - r}.

Exercise 2. Using the formula you proved in Exercise 1, calculate the value of the infinite series

4+27+149+1686+⋯4 + \frac{2}{7} + \frac{1}{49} + \frac{1}{686} + \cdots.

With the formula in exercise 1, it is an easy matter to see exactly when a geometric series converges or diverges, and its value in such a situation. If ∣r∣<1|r| < 1, then lim⁡n→∞rn=0\lim_{n \rightarrow \infty} r^{n} = 0, and if ∣r∣>1|r| > 1, then lim⁡n→∞rn\lim_{n \rightarrow \infty} r^{n} does not converge. Therefore, we have the following theorem:

Theorem. A geometric series ∑n=0∞arn=a+ar+ar2+…\sum_{n=0}^{\infty} ar^{n} = a + ar + ar^{2} + \ldots converges if ∣r∣<1|r| < 1 and diverges if ∣r∣>1|r| > 1. If ∣r∣<1|r| < 1, the value of the convergent geometric series is given by

a+ar+ar2+…=a1−r.a + ar + ar^{2} + \ldots = \frac{a}{1-r}.

A geometric series is a very, very special type of series. The fact that we have an explicit formula for the nnth partial sum, and that we can not only easily determine whether or not a geometric series converges or diverges, but also its value if it is convergent, is something which is quite rare. As you learn about various types of tests for determining whether or not series diverge or converge, always keep in the back of your mind these special facts about geometric series.

Exercise 3. What happens when the ratio of a geometric series is ±1\pm 1?

Exercise 4. Does the series

∑n=1∞n2n=12+24+38+416+532+…\sum_{n=1}^{\infty} \frac{n}{2^{n}} = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \frac{5}{32} + \ldots

converge or diverge? If it converges, what is its value?

Interactive · 8 questions

By now you've probably seen infinite series of some form, and you may even know how to evaluate some. First let's remind ourselves of the difference between series and sum.

Which of these is an infinite series?

Choose all correct answers

Read · 1 min

Sequence limits allow us to make sense of infinite sums, like the infinite geometric series 1+0.5+0.52+0.53+…1 + 0.5 + 0.5^2 + 0.5^3 + \dots. Namely, given an infinite sequence a0,a1,a2,a3,…a_0,a_1,a_2,a_3,\dots, the infinite sum (or infinite series) a0+a1+a2+a3+…a_0+a_1+a_2+a_3+\dots (also written ∑k=0∞ak\sum_{k=0}^{\infty} a_k sometimes) is defined to be the limit of partial sums: lim⁡n→∞(a0+⋯+an)\lim_{n\to\infty} (a_0+\dots+a_n) (or, written another way, lim⁡n→∞∑k=0nak\lim_{n\to\infty} \sum_{k=0}^{n} a_k). If the limit exists, the series is said to converge to the value of the limit; otherwise the limit diverges (possibly to infinity, possibly to nothing). For now we'll emphasize the algebra and intuition behind series manipulation, with examples of common series, but we'll do more with the limit definition later.

Type a concept, like “slope” or “photosynthesis”.

↑ ↓ to choose · Enter to openSearch explanations and problems