Wikipedia:Reference desk/Archives/Mathematics/2008 July 27

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July 27

Infinite Series

I have two questions about infinite series. The first one is a concept question. In the all the books I have read, it always says that if $a_{1},a_{2},a_{3},...$ are positive and decreasing ($a_{1}\geq a_{2}\geq a_{3}\geq ...\geq 0$), then the following alternating series $a_{1}-a_{2}+a_{3}-a_{4}+a_{5}-a_{6}+...$ converges if an only if the summands $a_{n}$ approach zero. My questions is that if all of these hypothesis are met, then $-a_{1}+a_{2}-a_{3}+a_{4}-a_{5}+a_{6}-...$ would also converge if and only if the summands approach zero as n goes to infinity? This is true right? Because all I am doing is that I am taking the first series, which is converging, and then multiplying it by -1 which will be well defined. And by a similar argument, $a_{2}-a_{3}+a_{4}-a_{5}+a_{6}-...$ will also converge if and only if the summands approach zero as n gets large, right? Because all I have done is multiplied by -1 and then added the first term?

The second question is, how do I find out in which region (in the upper half plane y>0) does the series $\sum _{n=1}^{\infty }(-1)^{n}{\frac {(\ln n)^{x}}{n^{y}}}$ diverges, converges conditionally, or converges absolutely?. Thanks!69.224.231.242 (talk) 07:33, 27 July 2008 (UTC)

First question - no. It's complicated - firstly the series a1,a2,a3 might not converge even if the terms tend to zero eg 1/1,1/2,1/3,1/4,1/5 etc

WRONG. You didn't read the question carefully. The answer is yes. Michael Hardy (talk) 18:05, 28 July 2008 (UTC)

Try reading the fucking question lord hardy - it doesn't make any sense because : quote "it always says that if $a_{1},a_{2},a_{3},...$ are positive and decreasing ($a_{1}\geq a_{2}\geq a_{3}\geq ...\geq 0$), then the following alternating ..."

I've underlined the two bits you need to check on. Hope that helps87.102.86.73 (talk) 20:41, 28 July 2008 (UTC)

Try reading that carefully and making any fucking sense out of it. B doesn't follow A.87.102.86.73 (talk) 20:40, 28 July 2008 (UTC)

Also try 1/2+1/3+1/4+1/5 etc does that converge - the numbers are decreasing??87.102.86.73 (talk) 20:49, 28 July 2008 (UTC)

Also check the bit about "if and only if the summands $a_{n}$ approach zero" - if a(n)=a(n) in the two series mentioned (rather than a1=a1-a2 a2=a3-a4 etc ) then the statement is true if "if and only if the summands $a_{n}$ approach zero". ie the part about the summands is irrelevent?87.102.86.73 (talk) 21:25, 28 July 2008 (UTC)

Second - the series a1,-a2,a3,-a4 etc - no . In fact there are series which oscillate and diverge which can be shown to have a real value (of the sum) at a limit of infinity.

Sorry you said 'summands' - if the 'summands' are a1-a2, a3-a4 etc then yes..

So the third question - no. obviously you are right that if the series A converges then the series -A also converges.

The problem is with your use of the term if and only if - if you remove that then your reasoning is correct. Ignore that - I missed the bit about summands.87.102.86.73 (talk) 12:06, 27 July 2008 (UTC)

You still need to bear in mind that a1>a2>a3>a4 etc does not guarantee convergence of the series.87.102.86.73 (talk) 12:08, 27 July 2008 (UTC)

Please see Convergent_series#Examples_of_convergent_and_divergent_series87.102.86.73 (talk) 12:11, 27 July 2008 (UTC)

I think you may find useful the two articles Alternating series and Alternating series test.87.102.86.73 (talk) 11:45, 27 July 2008 (UTC)

Have you tried combining odd and even terms into a new series and then testing for convergence using the 'ratio test' see Convergent_series#Convergence_tests.87.102.86.73 (talk) 11:31, 27 July 2008 (UTC) You would need to find the boundaries at which the series stops converging obviously.87.102.86.73 (talk) 11:44, 27 July 2008 (UTC)

Everything you say in the first paragraph is correct. For your second question, you'll probably want to use the alternating series test (that's what your first paragraph is about, if you haven't heard the name) for convergence. I haven't thought about absolute convergence yet. Algebraist 11:52, 27 July 2008 (UTC)

I agree. For absolute convergence, the hardest part is to work out the case of y=1 and figure out for which x this converges. After you do this, you get the answer for other y using the comparison test. For the tricky case y=1 (I would guess that there is a WP page on this, but I was not able to find it) you can use the integral test, if you know integration. If not, then it can also be done directly. Let us know and we can give you a few hints. Oded (talk) 16:42, 27 July 2008 (UTC)

The answer to the first question is yes. If

$a_{1}-a_{2}+a_{3}-a_{4}+\cdots \,$

converges, the so does

$-a_{1}+a_{2}-a_{3}+a_{4}-\cdots .\,$

Often the blind lead the blind on this page, it seems. Michael Hardy (talk) 18:03, 28 July 2008 (UTC)

You're not wrong there - because that is the answer to the SECOND question. As you say 'you didn't read the question carefully' well done.87.102.86.73 (talk) 21:06, 28 July 2008 (UTC)

Was I not clear enough? I already said 'Everything you say in the first paragraph is correct'. Algebraist 18:31, 28 July 2008 (UTC)

You were right, 87.102 was wrong. --Tango (talk) 18:35, 28 July 2008 (UTC)

Oh I was wrong - why didn't someone tell me.87.102.86.73 (talk) 20:36, 28 July 2008 (UTC)

I didn't because I could barely work out what you were talking about at all. Algebraist 21:25, 28 July 2008 (UTC)

Bit of politeness wouldn't kill everybody — whether you're pointing out an error or having one pointed out to you, try to keep it down to a dull roar, please, thanks guys! --_{tiny plastic} Grey Knight ⊖ 21:28, 28 July 2008 (UTC)

probability

probabiliti based short cut to solve question speedly —Preceding unsigned comment added by Anujay12 (talk • contribs) 18:42, 27 July 2008 (UTC)

What question? --Tango (talk) 18:55, 27 July 2008 (UTC)

This may be the biggest leap in question interpretation since "suitly emphazi", but I'm going to suggest Monte Carlo method as a way to use random input to help solve a problem. Confusing Manifestation(Say hi!) 23:19, 27 July 2008 (UTC)

Ah, it's been such a long time since I've "suitly emphazied" anything. :-) StuRat (talk) 17:31, 28 July 2008 (UTC)

Exponential Proportions

I was wondering, if i wanted to write 7^4 in cube form, what would it look like. I mean, how 64^1/2=2^3, what value of x makes 7^4=x^3. And how do you do that, so in the future i can do it myself. Simple terms please, just finished Algebra 1 last year.--Xtothe3rd (talk) 22:32, 27 July 2008 (UTC)

The value of x which makes $7^{4}=x^{3}$ is the cube root of $7^{4}$, which can be written as ${\sqrt[{3}]{7^{4}}}$ or $(7^{4})^{1/3}$. Using the rule $(a^{b})^{c}=a^{bc}$ it can be simplified to $7^{4/3}$. --tcsetattr (talk / contribs) 23:11, 27 July 2008 (UTC)

So then, (7^4)^(1 / 3) = 7^(4 / 3) = 13.3905183... more or less -hydnjo talk 01:45, 28 July 2008 (UTC)

You got be very careful.

If you write 7^4 = x^3

Then

x^3 = 2401

x = 13.39 or x = -6.69 - 11.6 i or x = -6.69 +11.6 i

or if you want exact answers

x = 7 (7)^(1/3) or x = -7 * (-7)^(1/3) or 7 * (-1)^(2/3) * (7)^(1/3)

122.107.182.1 (talk) 11:14, 28 July 2008 (UTC)