Calculus - Infinite Series

[rarisgod]

How would you go about finding the "sum" of this series:

1/[n(n+1)]

the sum is apparently 1...I don't recall having learned it in class but I could have forgotten with all of my other subjects piling up...so how would you go about this?

[karrot-x]

Take the limit as n goes to infinity, use l'hopital's.

EDIT: I'm wrong, looking for the correct answer.

[Thor]

Subbed.

I have totally forgotten all this stuff, so I may as
well get a quick refresher.

[wabbit]

I barely remember this stuff either but I can see that as n gets larger and approaches infinity, you are adding fractions to fractions and those fractions would just get smaller and smaller which would mean it would approach 1 but not actually get there .. I think.

[ehque]

The identity

1/[n(n+1)] = 1/n - 1/(n+1)

should help. Assuming you're talking about high school math (not college/university level math), splitting the function into individually actionable terms is the way to go. You can then use standard APGP math to work out your answer, or you can cancel out terms (as you can do with the expanded equation above).

Since the series of 1/n does not converge, you'll have to see if recurring terms can be cancelled out.

Sum to infinity for the series 1/[n(n+1)] = Sum to infinity for the series 1/n - 1/(n+1)

= (1/1 - 1/2) + ( 1/2 - 1/3 ) + (1/3 - 1/4 ) + ...

Since every fraction except 1/1 cancels out, simple observation will tell you this equation = 1. QED.

Originally Posted by wabbit

you are adding fractions to fractions and those fractions would just get smaller and smaller which would mean it would approach 1 but not actually get there ..

Half correct. It actually is mathematically equivalent to 1, because it is an infinite series.

Therefore, 1 = 1/2 + 1/4 + 1/8 + 1/16 + ... if the series is infinite. It is equal, not "nearly equal".

[Thor]

While the algebraic answer is correct, I think he was expecting to use calculus methodology, in this case derivatives, to examine the limit.

[Brad Barker]

two things to watch out for:

the harmonic series diverges as a logarithm http://en.wikipedia.org/wiki/Harmoni...mathematics%29

the series you posted isn't absolutely convergent. i.e., it has an answer, but it depends on how you sum the series.

ehque pwned this thread, btw. breaking up fractions like that is important. it's called "partial fraction decomposition." a week or so ago, i saw it in my grad-level solid state class (breaking up coulomb and bardeen-pines interaction terms), and i was like, "oh, yeah. forgot how to do that."

[Brad Barker]

Originally Posted by Thor

While the algebraic answer is correct, I think he was expecting to use calculus methodology, in this case derivatives, to examine the limit.

nope. dealing with infinite series is in the curriculum for calculus. in high school, it's a.p. calculus bc; in college it's a second semester course in calculus (often called "calculus II," though terminology differs).

most of the time, you really only deal with issues of convergence of infinite series; there are those dozen-and-a-half tests to perform. the really boring stuff. every so often, series can be analytically evaluated by students taking the course. many others have to wait. it's really an issue of being sufficiently clever, most of the time.

[Thor]

Cool. Well put. Thanks.

[stratovani]

This thread is giving me a headache!

[Ziltoid]

Currently finishing calculus II in college as well as my math lab ( honestly calc's are small peanuts aside the lab)

My two last math classes (changing tohuman sciences hahaha)

[Johnny Crab]

http://www.khanacademy.org/#Calculus

[rarisgod]

Sorry guys, yeah it's a second semester Calculus course, Integral Calculus (or Calculus II).

The sum is, indeed, 1. I just wasn't sure how to get to it. But now I realize that looking for canceling terms was discussed in class, I had just associated it with a later concept.

Also, does anyone know how to find the limit of e^(1/n)? My limit skills are weak and it's come back to haunt me with series...

[Disraeli Gears]

Originally Posted by rarisgod

Sorry guys, yeah it's a second semester Calculus course, Integral Calculus (or Calculus II).

The sum is, indeed, 1. I just wasn't sure how to get to it. But now I realize that looking for canceling terms was discussed in class, I had just associated it with a later concept.

Also, does anyone know how to find the limit of e^(1/n)? My limit skills are weak and it's come back to haunt me with series...

You're talking about the limit at infinity, right? I'm just in Calc I and I was terrible at series' in pre-calc, so I may be missing something here, but it seems to me as though since lim(n -> infinity) 1/n = 0 and x^0 = 1, lim(n -> infinity) e^(1/n) = 1.

[TortillaChip520]

Originally Posted by Thor

While the algebraic answer is correct, I think he was expecting to use calculus methodology, in this case derivatives, to examine the limit.

Nah, you don't take derivative for a series. You can use integral method to determine whether it converges or not, though.

What's sad is i'm learning this stuff right now, but i've given up on the class lol

[rarisgod]

What are the exponent laws for limits? Can you carry a limit into the exponent? If so, that's correct.

I know e is a constant, so that's obviously taken into account, but how do limits work with exponents? Ahh I hate limits...

[rarisgod]

Originally Posted by TortillaChip520

Nah, you don't take derivative for a series. You can use integral method to determine whether it converges or not, though.

What's sad is i'm learning this stuff right now, but i've given up on the class lol

Most of the people I know didn't make it in or dropped it already. There's only myself and one of my friends who decided to finish the course. I'm hoping I can make it for a pass

[Brad Barker]

Originally Posted by rarisgod

What are the exponent laws for limits? Can you carry a limit into the exponent? If so, that's correct.

I know e is a constant, so that's obviously taken into account, but how do limits work with exponents? Ahh I hate limits...

hmm, i sold my "advanced calculus" book which goes into the nitty-gritty with proving all of the various limit laws. the rule of thumb is that you can fiddle with limits as long as the limit in question isn't 0 or infinity. and indeed that is the case we're looking at.

perhaps replace 1/x with y. as x goes to infinity, y goes to 0.... i think that's fully justifiable at the level you're working with.

fun fact: the function exp(-1/x) shows up in superconductivity, which is a bit of a problem, since that function is so wonky.

[TortillaChip520]

Originally Posted by rarisgod

What are the exponent laws for limits? Can you carry a limit into the exponent? If so, that's correct.

I know e is a constant, so that's obviously taken into account, but how do limits work with exponents? Ahh I hate limits...

It depends on where the variable is...

If it's x squared over x, the limit is infinity, because as the numbers get bigger, a billion squared divided by a billion is still a huge number. If it's x over x squared, it approaches zero, because as numbers get bigger, a billion divided by a billion squared is going to to get smaller and smaller.

If it's 5x squared over 4x squared, it's going to approach 5/4, because the exponent on each variable is the same, so they're going to cancel out, and whatever the coefficients are will be the limit.

[Munjibunga]

Originally Posted by karrot-x

Take the limit as n goes to infinity, use l'hopital's.

EDIT: I'm wrong, looking for the correct answer.

I seem to remember l'hopital's rule said something like, if two functions are equal, so are there derivatives. Maybe I'm wrong.