idku
  • idku
I want to clarify something about the convergence of series and sequence.
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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idku
  • idku
So, for sequence to converge 1) Bound below and above 2) Monotonic (can be always decreasing or always increasing)
idku
  • idku
and for series: 3) The limit [n→ infinity] A(n) = 0 (am I correct)
zzr0ck3r
  • zzr0ck3r
yes, but the converse does not hold on 3)

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idku
  • idku
yes, what other conditions are there?
idku
  • idku
well, yes not on those 2 - harmonic series.
idku
  • idku
not those 3*
idku
  • idku
But, what else must be true?
zzr0ck3r
  • zzr0ck3r
of series?
idku
  • idku
yes convergence of series
zzr0ck3r
  • zzr0ck3r
there are many many theorems
idku
  • idku
Oh, like Σ 1/n^p then p>1 geometric series |r|<1 (and that is where ratio test is coming from) and others....
idku
  • idku
incl alternating series test.... and others
idku
  • idku
so it basically depends on a series in every unique cse?
idku
  • idku
case*
zzr0ck3r
  • zzr0ck3r
pretty much. there are some nice ways for some families of functions, but in general there is nothing that will deal with them all.
ganeshie8
  • ganeshie8
if you're comfortable with sequences, then you may think of any series as a sequence of partial sums
zzr0ck3r
  • zzr0ck3r
I also will assume we are talking about the real numbers with the regular topology.
idku
  • idku
yes, that is what i am reading on wiki and other sites ganeshie
idku
  • idku
yes. no imaginary sequences. I am not considering anything crazy
zzr0ck3r
  • zzr0ck3r
Getting 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
  • idku
alright
idku
  • idku
I was reading wiki just 3-5 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
  • idku
(i can create a new Q. if that is how the policy would play this time)
idku
  • idku
(sorry if i am typing to much, don't be afraid to tell me to shut up:))
zzr0ck3r
  • zzr0ck3r
This does not easily fall without other identities.
idku
  • idku
which identities?
zzr0ck3r
  • zzr0ck3r
series expansion of ln(x)
ganeshie8
  • ganeshie8
criterian for convergence of an alternating series is an interesting one i remember seeing a very nice proof from analysis
zzr0ck3r
  • zzr0ck3r
Maybe I am wrong, but I don't any basic ways to show this without more tools.
zzr0ck3r
  • zzr0ck3r
You can find the taylor expansion and plug in two...
zzr0ck3r
  • zzr0ck3r
But you are not proving anything....
idku
  • idku
wait can i start from taylor series of 1/x and integrate?
zzr0ck3r
  • zzr0ck3r
yeah
zzr0ck3r
  • zzr0ck3r
but you already know you can do that, so what is the point :)
zzr0ck3r
  • zzr0ck3r
unless you just feel like symbol pushing :)
ganeshie8
  • ganeshie8
\[\frac 1{1+x}=\sum_{n=0}^\infty (-x)^n\] integrate both sides and plugin \(x=1\) this is fun but yeah doesn't give you the bigger picture of beatiful alternating series
idku
  • idku
oh, we can get a series represenation for ln(x+1) :) nice
idku
  • idku
\(\large\color{black}{ \displaystyle \ln(x+1)=\sum_{ n=1 }^{ \infty } \frac{(-1)^nx^{n+1}}{n+1}}\)
ganeshie8
  • ganeshie8
ln(x) has no hope at x=0 but surely it behaves well for all finite positive values of x
idku
  • idku
oh yeah my index should have been n=0, i made that err
idku
  • idku
and nice to hear it works for other ln values
idku
  • idku
\(\large\color{black}{ \displaystyle \ln(x+1)=\sum_{ n=1 }^{ \infty } \frac{(-1)^nx^{n+1}}{n+1}}\) \(\large\color{black}{ \displaystyle \ln(1+1)=\sum_{ n=1 }^{ \infty } \frac{(-1)^n1^{n+1}}{n+1}}\)
idku
  • idku
oh, sorry n=0
idku
  • idku
\(\large\color{black}{ \displaystyle \ln(2)=\sum_{ n=0 }^{ \infty } \frac{(-1)^n}{n+1}}\)
ganeshie8
  • ganeshie8
now we're getting into radius of convergence and stuff for what values of x does that series converge ?
idku
  • idku
ok, let me do ratio test:)
idku
  • idku
|dw:1440308882162:dw|
idku
  • idku
oh, |x|<1
idku
  • idku
i left (-1)^n on purpose and hope correctly
ganeshie8
  • ganeshie8
right, so its kinda tricky, the domain is restricted to [-1, 1)
idku
  • idku
I 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
  • ganeshie8
this is one of the most celebrated series, there are like more than 100 different proofs for its convergence... let me dig them up... :)
idku
  • idku
we 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
  • idku
|dw:1440309328209:dw|this is kinda (a quite bad model) behavior of the partial sums of the alternating harmonic series
ganeshie8
  • ganeshie8
do you mean : r = \(\color{Red}{-1}\) is in the interval of convergence because we know that alternating harmonic series converges
idku
  • idku
oh, x=1, not r=1 i mean
idku
  • idku
all i was trying to say x=1 is valid, and i guess i typed much more unneccessary stuff on top
ganeshie8
  • ganeshie8
ahh okay you have that -1 in the front..
idku
  • idku
for interval of convergence?
ganeshie8
  • ganeshie8
\(\sum\limits_{n=0}^{\infty}\dfrac{x^n}{n}\) converges for all \(x\in [-1, 1)\)
idku
  • idku
yes, but when it is multiplied times (-1)^n then the interval is x= [-1,1] , isn't it?
ganeshie8
  • ganeshie8
right, we're on same page :)
idku
  • idku
oh, 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
  • idku
now that series makes sense ln(2) = 1 - 1/2 + 1/3 - 1/4 + 1/5 ...
idku
  • idku
thank you:)
idku
  • idku
well, should be (1)^(n\(\color{red}{-}\)1), but it is all sa,e
ganeshie8
  • ganeshie8
looks 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
  • idku
why is that better if I am permitted to plug in x=1 any way?
ganeshie8
  • ganeshie8
because when deriving that series representation, before integrating, we have used geometric series restricting ourselves to \((-1,1)\)
ganeshie8
  • ganeshie8
so technically we're not allowed to "plugin" x=1
idku
  • idku
oh, ok.... but we can always trick the math rules:) by the way you last line shouldn't have x^n on top, correct?
idku
  • idku
no nvm
idku
  • idku
well it is still same you are plugging in x=1, but you are technically not:O
ganeshie8
  • ganeshie8
limit of an infinite sum is not same as sum of limit
ganeshie8
  • ganeshie8
again, technically, we're not allowed to swap limit and infinite sum
ganeshie8
  • ganeshie8
thats the reason i left it as \(x\to 1\)
idku
  • idku
but 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
  • idku
the ratio test is 1, *if* x=1
idku
  • idku
maybe i can re-express myself
ganeshie8
  • ganeshie8
right, right, the variables are different so we can take the limit, no issues i guess
idku
  • idku
alrighty:) I guess then I am done for now
idku
  • idku
Will use a correct notation lim [n→∞] (thanks)^n
ganeshie8
  • ganeshie8
lookup the proof for convergence of alternating series, its really neat... im sure you will enjoy all other proofs of various convergence tests..
idku
  • idku
ok, lets see what they got;)
ganeshie8
  • ganeshie8
here 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
  • idku
oh:) 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)
idku
  • idku
gtg, sry bye
ganeshie8
  • ganeshie8
have good sleep :)

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