I want to clarify something about the convergence of series and sequence.

- idku

I want to clarify something about the convergence of series and sequence.

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- idku

So, for sequence to converge
1) Bound below and above
2) Monotonic (can be always decreasing or always increasing)

- idku

and for series:
3) The limit [n→ infinity] A(n) = 0
(am I correct)

- zzr0ck3r

yes, but the converse does not hold on 3)

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## More answers

- idku

yes, what other conditions are there?

- idku

well, yes not on those 2 - harmonic series.

- idku

not those 3*

- idku

But, what else must be true?

- zzr0ck3r

of series?

- idku

yes convergence of series

- zzr0ck3r

there are many many theorems

- 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

incl alternating series test.... and others

- idku

so it basically depends on a series in every unique cse?

- idku

case*

- 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

if you're comfortable with sequences, then you may think of any series as a sequence of partial sums

- zzr0ck3r

I also will assume we are talking about the real numbers with the regular topology.

- idku

yes, that is what i am reading on wiki and other sites ganeshie

- idku

yes. no imaginary sequences. I am not considering anything crazy

- 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

alright

- 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

(i can create a new Q. if that is how the policy would play this time)

- idku

(sorry if i am typing to much, don't be afraid to tell me to shut up:))

- zzr0ck3r

This does not easily fall without other identities.

- idku

which identities?

- zzr0ck3r

series expansion of ln(x)

- ganeshie8

criterian for convergence of an alternating series is an interesting one
i remember seeing a very nice proof from analysis

- zzr0ck3r

Maybe I am wrong, but I don't any basic ways to show this without more tools.

- zzr0ck3r

You can find the taylor expansion and plug in two...

- zzr0ck3r

But you are not proving anything....

- idku

wait can i start from taylor series of 1/x and integrate?

- zzr0ck3r

yeah

- zzr0ck3r

but you already know you can do that, so what is the point :)

- zzr0ck3r

unless you just feel like symbol pushing :)

- 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

oh, we can get a series represenation for ln(x+1) :) nice

- idku

\(\large\color{black}{ \displaystyle \ln(x+1)=\sum_{ n=1 }^{ \infty } \frac{(-1)^nx^{n+1}}{n+1}}\)

- ganeshie8

ln(x) has no hope at x=0
but surely it behaves well for all finite positive values of x

- idku

oh yeah my index should have been n=0, i made that err

- idku

and nice to hear it works for other ln values

- 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

oh, sorry n=0

- idku

\(\large\color{black}{ \displaystyle \ln(2)=\sum_{ n=0 }^{ \infty } \frac{(-1)^n}{n+1}}\)

- ganeshie8

now we're getting into radius of convergence and stuff
for what values of x does that series converge ?

- idku

ok, let me do ratio test:)

- idku

|dw:1440308882162:dw|

- idku

oh, |x|<1

- idku

i left (-1)^n on purpose and hope correctly

- ganeshie8

right, so its kinda tricky, the domain is restricted to [-1, 1)

- 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

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

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

|dw:1440309328209:dw|this is kinda (a quite bad model) behavior of the partial sums of the alternating harmonic series

- ganeshie8

do you mean :
r = \(\color{Red}{-1}\) is in the interval of convergence because we know that alternating harmonic series converges

- idku

oh, x=1, not r=1 i mean

- idku

all i was trying to say x=1 is valid, and i guess i typed much more unneccessary stuff on top

- ganeshie8

ahh okay you have that -1 in the front..

- idku

for interval of convergence?

- ganeshie8

\(\sum\limits_{n=0}^{\infty}\dfrac{x^n}{n}\) converges for all \(x\in [-1, 1)\)

- idku

yes, but when it is multiplied times (-1)^n
then the interval is x= [-1,1] , isn't it?

- ganeshie8

right, we're on same page :)

- 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

now that series makes sense
ln(2) = 1 - 1/2 + 1/3 - 1/4 + 1/5 ...

- idku

thank you:)

- idku

well, should be (1)^(n\(\color{red}{-}\)1), but it is all sa,e

- 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

why is that better if I am permitted to plug in x=1 any way?

- ganeshie8

because when deriving that series representation, before integrating, we have used geometric series restricting ourselves to \((-1,1)\)

- ganeshie8

so technically we're not allowed to "plugin" x=1

- 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

no nvm

- idku

well it is still same you are plugging in x=1, but you are technically not:O

- ganeshie8

limit of an infinite sum
is not same as
sum of limit

- ganeshie8

again, technically, we're not allowed to swap limit and infinite sum

- ganeshie8

thats the reason i left it as \(x\to 1\)

- 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

the ratio test is 1, *if* x=1

- idku

maybe i can re-express myself

- ganeshie8

right, right, the variables are different so we can take the limit, no issues i guess

- idku

alrighty:)
I guess then I am done for now

- idku

Will use a correct notation
lim [n→∞] (thanks)^n

- 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

ok, lets see what they got;)

- 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

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

gtg, sry bye

- ganeshie8

have good sleep :)

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