Find the sereis's radius and interval of convergence. For waht values of x does the series converge absolutely or conditionally?
\[\sum_{1}^{\infty}(4^nx^{2n})/n\]
well the lim of x^2n as n->inf = 0 when x < 1causing the sum to converge at some point since it will start adding 0+0+0...
By the ratio test, if the result of the following is less than 1, them series is guaranteed to converge:\[\lim_{n \rightarrow \infty}\left| \frac{4^{n+1}x^{2(n+1)}/(n+1)}{4^{n}x^{2n}/n} \right|=\lim_{n \rightarrow \infty}\left| \frac{4^{n+1}x^{2n+2}}{4^nx^{2n}}\frac{n}{n+1} \right|\]\[=\lim_{n \rightarrow \infty}\left| \frac{4.4^{n}x^{2n}x^{2}}{4^nx^{2n}}\frac{n}{n+1} \right|=\lim_{n \rightarrow \infty}\left| 4x^2\frac{1}{1+1/n} \right|\]\[=\left| 4x^2 \right|=4x^2\]So for convergence, we require,\[4x^2 \lt 1 \rightarrow -\frac{1}{2} \lt x \lt \frac{1}{2}\]
The radius of convergence is, by definition, all x that satisfy \[|x| < R\] where R is a non-negative, real number. \[|4x^2|=4|x^2|=4|x|^2<1 \rightarrow |x|<\frac{1}{2}\]
Since |x| < 1/2, the series is absolutely convergent for -1/2 < x < 1/2
Every absolutely convergent series is unconditionally convergent.
How about at end points? x=-1/2 and 1/2
I checked the end points, they both diverges, but the textbooks says it absolutely convergent for -1/2<x<1/2, and conditionally converges at x=-1/2
Good morning sir~
Yeah...it was late when I answered and hoped you wouldn't be considering the case where the magnitude of the ratio is equal to 1 (since anything can happen there).
haha morning
can you check the endpoints for me?
ok
When x= 1/2 (i.e. sub in 1/2) and simplify, you end up with the harmonic series, which is divergent. I'll do x=-1/2 now.
isn't it should be the same because x^(2n)
shouldn't it be *
Yeah, when x=-1/2, you get an alternating harmonic which I know to be equal to log(2) (natural log).
\[\frac{4^n \left( \frac{1}{2} \right)^{2n}}{n}=\frac{(2^2)^n \frac{1}{2^{2n}}}{n}=\frac{2^{2n}\frac{1}{2^{2n}}}{n}=\frac{1}{n}\]
yea i got this how about x=-1/2
Well, when you sub. x=-1/2, you get the same result as above, except for a (-1)^n outside the 1/n.
never mind i know where ive made a mistkae :)
This is the 'alternating harmonic series' which converges to ln(2). You can test convergence with the alternating series test.
so it conditionally converges at x=-1/2.. what i did i squared x , and then simplify the expresion so i always got 1/n :(
Actually, (-1/2)^(2n) is always positive (I'm eating while doing this so not completely concentrating :( )...so I need to have another look.
LOL alright (Lucas with -.-) i got to go to class now. see you soon haha
ok
Have u had another look?
No...sorry, I'm only back on because someone sent an e-mail. Is this something due now?
no i am just curious haha just get back to me whenever you can :)
If you have stuff to do, it is fine :]
one sec
It shouldn't be converging for either +1/2 or -1/2
Maybe the person who plugged in the answer for the book was easting and not paying attention like me before and ignored the '2' in 2n, and just focused on the 'n'.
*eating
oh, yea i tried..10 times yesterday,,,i still got the same answer,, but the book says it conditionally diverges at x=-1/2..
Book's wrong :)
Wolfram Alpha agrees
LOL my professor just told us the book is never wrong =-=
cuz he wrote part of the textbook we are using
haha, well, my professors wrote some books too and I have entire errata sheets handed out from them (i.e. corrections to their printed stuff-ups).
how embarrassing :P
we all make mistakes :)
It says inconclusive for both tests ..what does that mean?
It just says it used those two tests and found no answers...just like when we use tests and don't find answers and have to figure out another way.
If this thing converges, it will be ignoring the fact that (-1/2)^(2n) equals (-1)^(2n)(1/2)^(2n) = ((-1)^2)^n times (1/2)^(2n) = (1/2)^(2n)
Alright, so the book is WRONG..
I'm thinking...and if it's not, I'd like to know why.
Try a Cauchy Condensation test or something. Or alternating series test.
it is not an alternating series even if x is negative.. and i haven't learned cauchy condensation test :
Oh yea do you have time to explain the rearranging theorem ?
Yeah, exactly, that's my point. I don't think you can throw anything at this to prove convergence.
hmmm, probably not :(
I can later.
okiedokie
What's your question? At least I can think about it.
umm for a conditionally convergent series, you can rearrange the terms to get any sums you like?
the order of the terms
Sorry, I had to go do that. Some people on here are really starting to pellet me.
LOL i saw it ..that dude is retarded
So...anyways...yeah, I do it because I think sites like this are important and there are people on here (like you and a few others) who actually want to learn this stuff. The ones that piss me off are the ones crying for help on huge questions and then bugger off without even saying thank you.
Okay, so your question...I will scrounge a proof for you.
COOL buddy, don't waste your time and energy on aXX haha
The order of terms doesn't matter for a sum whose terms are positive: all rearrangements of a positive series converge to the same sum. It can make a difference if the series is alternating. For example, you can rearrange the terms of the alternating harmonic series to get 1/2 log 2 vs what it 'should' be (i.e. log 2).
It comes about when you start forming the partial sums: you can get different arrangements or patterns forming. This is okay if the sum is finite, but when you introduce infinity, it's a different game.
Because when you get your 'nth' partial sum and send n to infinity, you end up with different limits.
Is this what you're looking for?
sort of
The alternating series is a conditionally-convergent series, and it's also a series where you can get different sums depending on the rearrangements...so, yes, you can get different sums in a conditionally convergent series.
do you mean the pattern of the series changes if we change the order of the terms, so as n->infinity, the limit is different?
Exactly
oh i feel you ^^ thank you ~~~~~
I'll send something I have. It shows how you can arrange the alternating harmonic to get 1/2 log 2 instead of log 2.
wait how about asolutely convergent series, if we rearrange the oders of terms, the pattern of the series also changes right?
All the terms in an abs. conv. series are positive, so as I said before, they will sum to the same limit no matter the arrangement. It's the introduction of two things that stuffs infinite sums up: 1) minus sign 2) infinity
This is crazy lol
Has your professor shown you why an infinite series of positive terms doesn't change it's convergence (sum) upon rearrangement?
nope he didn't get into any details, he just stated the theorem and said this is not in the scope of this course..
oh...
The proof is on that sheet too.
yea i am looking at it
ok...i'll leave it with you. I have to go do some work. Have fun with it.
thanks !!! have fun with your work too
bye :)
c ya:)
LUCAS~~~ I have a question regarding PHYSICS~~
k
a wooden block is at rest in a vehicle that is also at rest, if the vehicle accelerates north, why will the block slide southward
one sec
okie okie
inertia
sorry, ordered chinese food and guy was at door
first law of motion
oh yea chinese food haha~
A 34.0kg packing case is initially at rest on the floor of a 1380kg pickup truck. The coefficient of static friction between the case and the truck floor is 0.30, and the coefficient of kinetic friction is 0.20. Before each acceleration given below, the truck is traveling due north at constant speed. a1=2 m/s^2
what's the question?
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