- 19 Mar '09 05:14 / 1 edit

You can kinda see that as n goes to infinity a(n) goes to 2. However, I think you worded the problem incorrectly. You use n as the variable and then refer to the nth term.*Originally posted by joe shmo***having trouble defining the nth term in the following series**

the starting series is of the form

a(n) = 1/n

and I want to determine the sum (1/n) from n = 1 to infinity

so

a(1) = 1

a(2) = 3/2

a(3)= 11/6

a(4) = 25/12

so can anyone help me out on defining the nth term?

I'm having a little trouble figuring out how exactly to do this. You might want to look here:

http://www.math.toronto.edu/mathnet/plain/questionCorner/geomsum.html

I'm going to use x as my variable.

Sum a(x) as x goes from 1 to infinity = 1 + 1/2 + 1/3 + ... + 1/n + ... + 0.

The nth term is 1/n. Right?

Finding the sum is a different problem. Right? - 19 Mar '09 05:16

there is no real easy way to define sum[a(n)] for the n'th term... the common denominator of the fractions would be the product of all primes less than or equal to n, and the numerators would all be a mess.*Originally posted by joe shmo***having trouble defining the nth term in the following series**

the starting series is of the form

a(n) = 1/n

and I want to determine the sum (1/n) from n = 1 to infinity

so

a(1) = 1

a(2) = 3/2

a(3)= 11/6

a(4) = 25/12

so can anyone help me out on defining the nth term?

this is a pretty famous series called the "harmonic series" and it can be proven that as n -> infinity, a(n) also approaches infinity. it's the classic example of showing that even if the individual terms of the sequence get arbitrarily small, the series can diverge to infinity.

i won't prove it for you here because it's pretty magical to see for the first time yourself, but try looking at the series term by term and showing that each term is greater than or equal to the corresponding terms of a different series that you KNOW diverges to infinity. good luck, and if you want to see it just post asking for the proof. - 19 Mar '09 05:19 / 1 edit

note: the sum of all 1/n as n-> infinity is VERY different from the sum 1/n^2 as n-> infinity. in fact, (as you can tell from ATY's link) 1/n^2 converges to a finite sum as n goes to infinity! as the proof shows, it is always smaller than 2. however, the harmonic series has no least upper bound - if you pick an arbitrarily large number, and go far enough out into the partial sums, the series 1/n will trump whatever number you picked and continue growing.*Originally posted by AThousandYoung***You can kinda see that as n goes to infinity a(n) goes to 2. However, I think you worded the problem incorrectly. You use n as the variable and then refer to the nth term.**

I'm having a little trouble figuring out how exactly to do this. You might want to look here:

http://www.math.toronto.edu/mathnet/plain/questionCorner/geomsum.html

if you continue through calc 2, you'll learn about the "power test" for convergence and divergence of series, as well as a number of other techniques for proving convergence or divergence (i.e. the "comparison test," the "improper integral test," etc.). it's a really elegant topic for those so inclined to explore! - 19 Mar '09 05:20 / 1 edit

I've gone well beyond Calc 2 but I probably didn't absorb everything.*Originally posted by Aetherael***note: the sum of all 1/n as n-> infinity is VERY different from the sum 1/n^2 as n-> infinity. if you continue through calc 2, you'll learn about the "power test" for convergence and divergence of series, as well as a number of other techniques for proving convergence or divergence.** - 19 Mar '09 06:13Another interesting way to look at this type of thing is to ask the question "Given an infinite set of positive integers, how infrequently must they occur (compared to all the positive integers) in order that the sum of their reciprocals converge?" For example, the set of all integers is, as this thread shows, too "dense" in this sense, but the set of perfect squares is not.

This criterion is interesting because it is more discriminating than some simpler measures of the "size" of an infinite set of integers. For example, there is a meaningful sense in which the "probability" that a randomly chosen integer is prime is 0. However, you can check (although it is not easy) that SUM(1/p : p prime) diverges. - 19 Mar '09 10:12

It's an abuse of language* to say "randomly chosen integer".*Originally posted by ChronicLeaky***Another interesting way to look at this type of thing is to ask the question "Given an infinite set of positive integers, how infrequently must they occur (compared to all the positive integers) in order that the sum of their reciprocals converge?" For example, the set of all integers is, as this thread shows, too "dense" in this sense, but the set of ...[text shortened]... s 0. However, you can check (although it is not easy) that SUM(1/p : p prime) diverges.**

There is no such thing as an uniform distribution over an unbounded set, like the integers, so you cannot randomly generate an integer from a distribution where each integer is equally likely.

*and maybe this is why you wrote probability between quotes, but I still felt this had to be qualified. - 19 Mar '09 18:07

Quite right, it is an abuse of language. I used the word "probability" because people are, I think, more likely to know what I mean than if I had said "Schnirelmann density" or something.*Originally posted by Palynka***It's an abuse of language* to say "randomly chosen integer".**

There is no such thing as an uniform distribution over an unbounded set, like the integers, so you cannot randomly generate an integer from a distribution where each integer is equally likely.

*and maybe this is why you wrote probability between quotes, but I still felt this had to be qualified. - 20 Mar '09 03:39 / 1 editthanks to all for the insightful information, As for the proof ( Aetherael) I don't think I'm ready yet, I will have to come back to this idea after I have completed some more study the subject...P.s. you hit the nail on the head, I am a calc 2 student seeing this stuff for the first time. The theory behind infinte series captured my attention some time ago, and I am now finally getting to analyze a small portion of the ideas....its pretty interesting stuff!
- 20 Mar '09 08:46

this is a great topic, but definitely very difficult to grasp the first time around - keep at it! this series is a classic caveat that just because the individual terms of an infinite series approach zero (which is a prerequisite for the series converging to a finite sum), it does not necessarily mean that the series DOES converge! more investigation is required.*Originally posted by joe shmo***thanks to all for the insightful information, As for the proof ( Aetherael) I don't think I'm ready yet, I will have to come back to this idea after I have completed some more study the subject...P.s. you hit the nail on the head, I am a calc 2 student seeing this stuff for the first time. The theory behind infinte series captured my attention some time ag ...[text shortened]... now finally getting to analyze a small portion of the ideas....its pretty interesting stuff!**

you will spend a good portion of the rest of this topic in calc 2 comparing other series to sigma(1/n) and sigma(1/n^2) in order to prove their divergence or convergence, respectively. it's great fun once you get a handle on it.