From Encyclopedia

A sequence is an ordered listing of numbers such as {1, 3, 5, 7, 9}. In mathematical terms, a sequence is a function whose domain is the natural number set {1, 2, 3, 4, 5, â€¦}, or some subset of the natural numbers, and whose range is the set of real numbers or some subset thereof. Let f denote the sequence {1, 3, 5, 7, 9}. Then f (1) = 1, f (2) = 3, f (3) = 5, f (4) = 7, and f (5) = 9. Here the domain of f is {1, 2, 3, 4, 5} and the range is {1, 3, 5, 7, 9}. The terms of a sequence are often designated by subscripted variables. So for the sequence {1, 3, 5, 7, 9}, one could write a 1 = 1, a 2 = 3, a 3 = 5, a 4 = 7, and a 5 = 9. A shorthand designation for a sequence written in this way is {a n}. It is sometimes possible to get an explicit formula for the general term of a sequence. For example, the general nth term of the sequence {1, 3, 5, 7, 9} may be written a n = 2n âˆ’ 1, where n takes on values from the set {1, 2, 3, 4, 5}. The sequence {1, 3, 5, 7, 9} is finite, since it contains only five terms, but in mathematics, much work is devoted to the study of infinite sequences.* For instance, the sequence of "all" odd natural numbers is written {1, 3, 5, 7, 9, â€¦}, where the three dots indicate that there is an infinite number of terms following 9. Note that the general term of this sequence is also a n = 2n âˆ’ 1, but now n can take on any natural number value, however large. Other examples of infinite sequences include the even natural numbers, all multiples of 3, the digits of pi (Ï€), and the set of all prime numbers. *A famous infinite sequence is the so-called Fibonacci sequence {1, 1, 2, 3, 5, 8, 13, 21, â€¦} in which the first two terms are each 1 and every term after the second is the sum of the two terms immediately preceding it. The amount of money in a bank account paying compound interest at regular intervals is a sequence. If $100 is deposited at an annual interest rate of 5 percent compounded annually, then the amounts of money in the account at the end of each year form a sequence whose general term can be computed by the formula a n = 100(1.05)n, where n is the number of years since the money was deposited. A series is just the sum of the terms of a sequence. Thus {1, 3, 5, 7, 9} is a sequence, but 1 + 3 + 5 + 7 + 9 is a series. So long as the series is finite the sum may be found by adding all the terms, so 1 + 3 + 5 + 7 + 9 = 25. If the series is infinite, then it is not possible to add all the terms by the ordinary addition algorithm , since one could never complete the task. Nevertheless, there are infinite series that have finite sums. An example of one such series is: â€¦, where, again, the three dots indicate that this series continues according to this pattern without ever ending. Now, obviously someone cannot sit down and complete the term by term addition of this series, even in a lifetime, but mathematicians reason in the following way. The sum of the first term is 1; the sum of the first two terms is 1.5; the sum of the first three terms is 1.75; the sum of the first four terms is 1.875; the sum of the first five terms is 1.9375; and so on. These are called the first five partial sums . Mathematically, it is possible to argue that no matter how many partial sums one takes, the n th partial sum will always be slightly less than 2 (in the above example), for any value of n. On the other hand, each new partial sum will be closer to 2 than the one before it. Therefore, one would say that the limiting value of the sequence of partial sums is 2 and the sum is defined as the infinite series â€¦. This infinite series is saidto converge to 2. A case where the partial sums grow without bound is described as a series that diverges or that has no finite sum. see also Fibonacci, Leonardo Pisano. Stephen Robinson Narins, Brigham, ed. World of Mathematics. Detroit: Gale Group, 2001.

From Omilili

### Partial Sumations of the **Divergent** Prime Reciprocal Series

Concerning this function... f(n) = \sum_{i=**1**}^{n}\frac{

**1**}{p_{i}} ...where... p_{i} = i\, th doesn't converge. Just now I graphed f(n) for values of n between

**1**and 20,000 and the results surprised ://mathworld.wolfram.com/HarmonicSeriesofPrimes.html $$\sum_{p \in primes}^

**x**1/p = \ln\ln

**x**+ B_1 + o(

**1**...

### Simple? **Mathematical****formula** for two pots connected in series and parallel...

To be a simple task - I'm trying to work out a **formula**to represent the relationship between the two pots / parallel sums but can't seem to get the correct

**formula**to calculate the total resistance at points Hi there, this is the first time I have posted on this forum, I have what would appear to be a simple task - I'm trying to...

### A question about sum of **sequences**

Hello! I am droddling with the harmonic **sequence**. 1. I know this: S 1 = 1/1 + 1/2 + 1/3 + 1/4 + 1/5 can't willy-nilly rearrange terms in a conditionally

**convergent**series like you did. Homepage: http

**convergent**(value already given.) S3 is divergent for the same reason S1 is. And, although S4 is a term

From Yahoo Answers

**Question:**I'm currently a Sophomore in high school and i'm taking Algebra 2. As hard as I try, I cannot remember having ever learned about what a "geometric series" is in past Algebra 1 and Geometry classes. Can anyone help me with this problem? It says to solve for the "sum of a geometric series": S= a(1-r^n) \ 1- r S = 80 r = 3 n = 4

**Answers:**S = a(1 - r^n) / (1 - r) is one of the formulas for the sum of a geometric series You've been given the sum (S), common ratio (r) and number of terms (n) so I guess they want you to find the first term of the series (a) r > 1 so it's probably better to use the other form of the formula for the sum: S = a(r^n - 1) / (r - 1) subsing what you've been given: 80 = a(3^4 - 1) / (3 - 1) 80 = 80a / 2 80 = 40a a = 2 So the first term of the geometric series is 2

**Question:**I'm studying for a test and doing practice papers, and I just don't know how they got the answer. The question is: The sequence, 5, 7, 11 ... has as its general term 2^n+3 Find a) the 4th term (which I did and it's 19) b) a formula for the sum of the first n terms. I'm stuck on b. The answer they have is: 2(2^n-1)+3n How did they get that???? Thank you thank you (Ps this is a geometric series right?)

**Answers:**Actually no, it's not a geometric series because each term is 2^n + 3, not in the form of a^r. However, 2^n itself is a geometric series, and we can make use of that here. Remember that you can always split up (a+b) into ( a) + ( b). So split this up into ( 2^k) + ( 3), where k goes from 1 to n. The first term IS a geometric series where a=1 and r=2. You know how to write the partial sum for that; it's just a(r^n - 1)/(r - 1). As for the " 3" term, well that's just "3 + 3 + 3 + ... + 3" for a total of n times, so the sum is just 3n. Now add up both expressions to get the answer, 2(2^n - 1) + 3n.

**Question:**Determine whether the series is absolutely convergent, conditionally convergent, or divergent. the sum from n=1 to 2 (-1)^n * (3)/(n+4^n)

**Answers:**For this problem you can use either root test or ratio test. The ratio test in this case is easier. The ratio test is: lim n->infinity |a_n+1 / a_n| (for the sake a clarity assume everything is absolute | |) = lim n->infinity [(-1)^(n+1)*(3) / (n+1)+4^n+1 ] * [(n)+4^n)/(-1)^n3] = lim n->infinity n/n+5 = 1 1 , is inconclusive.. so we need to use another test. Using the test for alternating series.. when... lim n->infinity bn bn+1 < bn the series is convergent. bn = 3/(n+4^n ) lim n -> infinity 3/(n+4^n) = 0 ..Therefore, first part of test is true. b2 = 0.1666 b1 = 0.6 ...Therefore b2 < b2 , and second part of test is true. And so, we know that it is convergent. However, we also have to test for -3n/(n+4^n). We can use the comparison test, which states that if lim n-> infinity an/bn =c, and c > 0 then both series converge or both series diverge. we can compare with the already known divergent harmonic series 1/n..just tweaked.. an = -3n/n+4^n bn = -3n/4^n lim n-> infinity = an/bn lim n->infinity = 4^n / n+4^n ...= 1 therefore, because c > 1, since bn is divergent an is also divergent. and so, the series is conditionally convergent:)

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