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idku
 one year ago
I want to clarify something about the convergence of series and sequence.
idku
 one year ago
I want to clarify something about the convergence of series and sequence.

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idku
 one year ago
Best ResponseYou've already chosen the best response.2So, for sequence to converge 1) Bound below and above 2) Monotonic (can be always decreasing or always increasing)

idku
 one year ago
Best ResponseYou've already chosen the best response.2and for series: 3) The limit [n→ infinity] A(n) = 0 (am I correct)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1yes, but the converse does not hold on 3)

idku
 one year ago
Best ResponseYou've already chosen the best response.2yes, what other conditions are there?

idku
 one year ago
Best ResponseYou've already chosen the best response.2well, yes not on those 2  harmonic series.

idku
 one year ago
Best ResponseYou've already chosen the best response.2But, what else must be true?

idku
 one year ago
Best ResponseYou've already chosen the best response.2yes convergence of series

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1there are many many theorems

idku
 one year ago
Best ResponseYou've already chosen the best response.2Oh, like Σ 1/n^p then p>1 geometric series r<1 (and that is where ratio test is coming from) and others....

idku
 one year ago
Best ResponseYou've already chosen the best response.2incl alternating series test.... and others

idku
 one year ago
Best ResponseYou've already chosen the best response.2so it basically depends on a series in every unique cse?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1pretty much. there are some nice ways for some families of functions, but in general there is nothing that will deal with them all.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2if you're comfortable with sequences, then you may think of any series as a sequence of partial sums

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1I also will assume we are talking about the real numbers with the regular topology.

idku
 one year ago
Best ResponseYou've already chosen the best response.2yes, that is what i am reading on wiki and other sites ganeshie

idku
 one year ago
Best ResponseYou've already chosen the best response.2yes. no imaginary sequences. I am not considering anything crazy

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Getting good with sequences is a great idea if you are planning on doing analysis. There are definitions of continuity that uses sequences that make things much nicer than dealing with the standard \(\epsilon \delta\) definition

idku
 one year ago
Best ResponseYou've already chosen the best response.2I was reading wiki just 35 minutes ago, and I saw that \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } \frac{(1)^{n+1}}{n}=\ln(2)}\) (they don't post any link to an explanation)

idku
 one year ago
Best ResponseYou've already chosen the best response.2(i can create a new Q. if that is how the policy would play this time)

idku
 one year ago
Best ResponseYou've already chosen the best response.2(sorry if i am typing to much, don't be afraid to tell me to shut up:))

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1This does not easily fall without other identities.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1series expansion of ln(x)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2criterian for convergence of an alternating series is an interesting one i remember seeing a very nice proof from analysis

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Maybe I am wrong, but I don't any basic ways to show this without more tools.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1You can find the taylor expansion and plug in two...

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1But you are not proving anything....

idku
 one year ago
Best ResponseYou've already chosen the best response.2wait can i start from taylor series of 1/x and integrate?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1but you already know you can do that, so what is the point :)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1unless you just feel like symbol pushing :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac 1{1+x}=\sum_{n=0}^\infty (x)^n\] integrate both sides and plugin \(x=1\) this is fun but yeah doesn't give you the bigger picture of beatiful alternating series

idku
 one year ago
Best ResponseYou've already chosen the best response.2oh, we can get a series represenation for ln(x+1) :) nice

idku
 one year ago
Best ResponseYou've already chosen the best response.2\(\large\color{black}{ \displaystyle \ln(x+1)=\sum_{ n=1 }^{ \infty } \frac{(1)^nx^{n+1}}{n+1}}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2ln(x) has no hope at x=0 but surely it behaves well for all finite positive values of x

idku
 one year ago
Best ResponseYou've already chosen the best response.2oh yeah my index should have been n=0, i made that err

idku
 one year ago
Best ResponseYou've already chosen the best response.2and nice to hear it works for other ln values

idku
 one year ago
Best ResponseYou've already chosen the best response.2\(\large\color{black}{ \displaystyle \ln(x+1)=\sum_{ n=1 }^{ \infty } \frac{(1)^nx^{n+1}}{n+1}}\) \(\large\color{black}{ \displaystyle \ln(1+1)=\sum_{ n=1 }^{ \infty } \frac{(1)^n1^{n+1}}{n+1}}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.2\(\large\color{black}{ \displaystyle \ln(2)=\sum_{ n=0 }^{ \infty } \frac{(1)^n}{n+1}}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2now we're getting into radius of convergence and stuff for what values of x does that series converge ?

idku
 one year ago
Best ResponseYou've already chosen the best response.2ok, let me do ratio test:)

idku
 one year ago
Best ResponseYou've already chosen the best response.2i left (1)^n on purpose and hope correctly

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2right, so its kinda tricky, the domain is restricted to [1, 1)

idku
 one year ago
Best ResponseYou've already chosen the best response.2I will dare to sya in a face of god you are wrong (apologize in advance) I was just about say \(\large\color{black}{ \displaystyle \ln(1+1)=\sum_{ n=0 }^{ \infty } \frac{(1)^n1^{n+1}}{n+1}}\) \(\large\color{black}{ \displaystyle \ln(1+1)=\sum_{ n=0 }^{ \infty } \frac{(1)^n1^{n+1}}{n+1}}\) \(\large\color{black}{ \displaystyle \ln(2)=\sum_{ n=0 }^{ \infty } \frac{(1)^n}{n+1}}\) and because i know alternative harmonic converges, thus r=1 is valid in this case

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2this is one of the most celebrated series, there are like more than 100 different proofs for its convergence... let me dig them up... :)

idku
 one year ago
Best ResponseYou've already chosen the best response.2we can make r=1, and just because if x=1 we get a result of "1" from the ratio test it doesn't tell us that series diverges, it is inconclusive when limit comes out with r=1, the domain is including 1.

idku
 one year ago
Best ResponseYou've already chosen the best response.2dw:1440309328209:dwthis is kinda (a quite bad model) behavior of the partial sums of the alternating harmonic series

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2do you mean : r = \(\color{Red}{1}\) is in the interval of convergence because we know that alternating harmonic series converges

idku
 one year ago
Best ResponseYou've already chosen the best response.2all i was trying to say x=1 is valid, and i guess i typed much more unneccessary stuff on top

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2ahh okay you have that 1 in the front..

idku
 one year ago
Best ResponseYou've already chosen the best response.2for interval of convergence?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\(\sum\limits_{n=0}^{\infty}\dfrac{x^n}{n}\) converges for all \(x\in [1, 1)\)

idku
 one year ago
Best ResponseYou've already chosen the best response.2yes, but when it is multiplied times (1)^n then the interval is x= [1,1] , isn't it?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2right, we're on same page :)

idku
 one year ago
Best ResponseYou've already chosen the best response.2oh, and then we extract that representation \(\large\color{black}{ \displaystyle \ln(\color{red}{1}+1)=\sum_{ n=0 }^{ \infty } \frac{(1)^n\color{red}{1}^{n+1}}{n+1}}\) \(\large\color{black}{ \displaystyle \ln(2)=\sum_{ n=0 }^{ \infty } \frac{(1)^n}{n+1}}\) \(\large\color{black}{ \displaystyle \ln(2)=\sum_{ n=1 }^{ \infty } \frac{(1)^{n+1}}{n}}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.2now that series makes sense ln(2) = 1  1/2 + 1/3  1/4 + 1/5 ...

idku
 one year ago
Best ResponseYou've already chosen the best response.2well, should be (1)^(n\(\color{red}{}\)1), but it is all sa,e

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2looks good! its not a big deal, but below looks better : \(\large\color{black}{ \displaystyle \ln(\color{red}{x}+1)=\sum_{ n=0 }^{ \infty } \frac{(1)^n\color{red}{x}^{n+1}}{n+1}}\) \(\large\color{black}{ \displaystyle \lim\limits_{x\to 1}~\ln(\color{red}{x}+1)=\lim\limits_{x\to 1}~\sum_{ n=0 }^{ \infty } \frac{(1)^n}{n+1}}\) \(\large\color{black}{ \displaystyle \ln(2)=\lim\limits_{x\to 1}~\sum_{ n=1 }^{ \infty } \frac{(1)^{n}x^n}{n}}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.2why is that better if I am permitted to plug in x=1 any way?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2because when deriving that series representation, before integrating, we have used geometric series restricting ourselves to \((1,1)\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2so technically we're not allowed to "plugin" x=1

idku
 one year ago
Best ResponseYou've already chosen the best response.2oh, ok.... but we can always trick the math rules:) by the way you last line shouldn't have x^n on top, correct?

idku
 one year ago
Best ResponseYou've already chosen the best response.2well it is still same you are plugging in x=1, but you are technically not:O

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2limit of an infinite sum is not same as sum of limit

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2again, technically, we're not allowed to swap limit and infinite sum

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2thats the reason i left it as \(x\to 1\)

idku
 one year ago
Best ResponseYou've already chosen the best response.2but plugging x=1 does not make the series diverge tho' the ratio test is 1 is x=1 (and jsut that no conclusions can be made, but not that it will make the sum diverge). THis series is conditionally convergent when x=1. I don't really see much problem with actually going ahead and plugging x=1. It is just illogical to me:( :)

idku
 one year ago
Best ResponseYou've already chosen the best response.2the ratio test is 1, *if* x=1

idku
 one year ago
Best ResponseYou've already chosen the best response.2maybe i can reexpress myself

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2right, right, the variables are different so we can take the limit, no issues i guess

idku
 one year ago
Best ResponseYou've already chosen the best response.2alrighty:) I guess then I am done for now

idku
 one year ago
Best ResponseYou've already chosen the best response.2Will use a correct notation lim [n→∞] (thanks)^n

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2lookup the proof for convergence of alternating series, its really neat... im sure you will enjoy all other proofs of various convergence tests..

idku
 one year ago
Best ResponseYou've already chosen the best response.2ok, lets see what they got;)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2here it is https://www.youtube.com/watch?v=tt430qiyIds&index=19&list=PL04BA7A9EB907EDAF just watch it like a movie, don't wry about all the weird terms that you might hear... :)

idku
 one year ago
Best ResponseYou've already chosen the best response.2oh:) It's 2:30 am in my location, i was sitting late at night because I was curious... this movie is too long for now. i have to go offline right now:) I saved the link (and if anything i can login to see the link)..... thank you very much! C(U)
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