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whooping_a_panda
04-07-05, 02:10 AM
help!!! I'm terribly and utterly lost and confused. Okay well maybe you can't help me with that but perhaps a little guidance on my calc would suffice.

In the infinite series(-1)^k/((k+1)(k+2))^.5 can i say its absolute series diverges based on the p series argument? since p < 1 or does the fact that i have a polynomial down their complicate matters and if so how do i fix it?

Frodo Baggins
04-07-05, 06:36 AM
help!!! I'm terribly and utterly lost and confused. Okay well maybe you can't help me with that but perhaps a little guidance on my calc would suffice.

In the infinite series(-1)^k/((k+1)(k+2))^.5 can i say its absolute series diverges based on the p series argument? since p < 1 or does the fact that i have a polynomial down their complicate matters and if so how do i fix it?

No, you cannot, as it's not in the form of a 'p-series', at least not right away.

Well it's easy to see absolute divergence. For large values of k, the absolute value of the series does behave like 1/sqrt(k^2), which diverges.

The easy way to do this is to do a comparison:
1/sqrt(k^2 + 3k + 2) >= 1/sqrt(k^2 + 3k^2 + 2k^2) = 1/sqrt(6k^2) which diverges (harmonic series).

Conditional convergence follows, of course, as the general term tends to zero and is monotonically decreasing.

threeme2189
04-07-05, 09:48 AM
im completely lost.

Slackfumasta
04-07-05, 09:59 AM
*drool*

I'm almost thirty years old, and I just learned a way to tie my shoes so that they don't come untied.

I'm serious.

whooping_a_panda
04-07-05, 12:11 PM
thank you frodo. That makes a lot of sense, dont know why I couldnt think of it earlier.

Frodo Baggins
04-07-05, 02:18 PM
The general trick to use for most series is to observe what happens at infinity. Generally, this simplifies the general term. Then use a comparison to get what you want.

whooping_a_panda
04-07-05, 06:48 PM
hmmm, i believe all physics majors should be paired with a math major. That way so much more would get done with much less estimation.

Now can you prove the gamma function to be an irrational number? ;) I'll mail you a box of chips a'hoy if you do. :D

Frodo Baggins
04-07-05, 08:38 PM
Now can you prove the gamma function to be an irrational number? ;) I'll mail you a box of chips a'hoy if you do. :D

Hmm?
The gamma function is not a number. It's the definite integral of t^(z-1)e^-t evaluated from 0 to infinity for z complex with Re(z) > 0.

I think you're referring to 'gamma', which is often used to represent the Euler-Mascheroni constant, which is the limit of the difference between the n'th partial sum of the harmonic series and the natural logarithm:

lim_(n to infinity) { Sum_{k=1}^n 1/k - ln(n) }

It is not known if this constant is irrational, let alone transcendental (Wells 1986, p. 28). The famous English mathematician G. H. Hardy is alleged to have offered to give up his chair at Oxford to anyone who proved [gamma] to be irrational (Havil 2003, p. 52), although no written reference for this quote seems to be known. Hilbert mentioned the irrationality of [gamma] as an unsolved problem that seems "unapproachable" and in front of which mathematicians stand helpless (Havil 2003, p. 97). Conway and Guy (1996) are "prepared to bet that it is transcendental," although they do not expect a proof to be achieved within their lifetimes.

Interesting, though the words "Who cares?" springs to mind. Personally, I'm not going to lose any sleep not knowing that the difference between the harmonic series and natural logarithm is rational, irrational, or transcendental.

whooping_a_panda
04-07-05, 09:46 PM
Interesting, though the words "Who cares?" springs to mind.

Glad it came from a math guy, well the box of chips a'hoy is on the table should anyone have the answer to this. :beer:

Frodo Baggins
04-07-05, 10:24 PM
Glad it came from a math guy, well the box of chips a'hoy is on the table should anyone have the answer to this. :beer:

A math professor once told me: Mathematicians are people who are able to suspend all sense of reality for weeks at a time.

Malpine Walis
04-07-05, 11:40 PM

whooping_a_panda
04-08-05, 01:40 AM
okay Im at it again, with a few more questions.

1) i have the series k(3/4)^k - can I say this is a converging sereis because it is a geometric sereis with abs(r) which is abs(3/4) < 1 ?

2) this one i have no clue on. the sereis k^e/e^k - and actually if you could explain to me what happens at any value say x is devided by a constant^k or c^k so x/c^k( i imagine again convergence or divergence is based on the geometric sereis test but im not sure.

there may be more to follow, but well see.

*note dont worry this isnt something i will be turning in this is for my own benefit so i dont fall too badly on the exam.

Frodo Baggins
04-08-05, 08:13 AM
2) this one i have no clue on. the sereis k^e/e^k - and actually if you could explain to me what happens at any value say x is devided by a constant^k or c^k so x/c^k( i imagine again convergence or divergence is based on the geometric sereis test but im not sure.

Why do you think it's supposed to use the geometric series? (No really, I want to know...)

Apply the root test (take the k'th root) and examine the limit:
lim_{k to infinity} k^(e/k)/e = ??

Disregard the 1/e constant. The limit k^(e/k) approaches 1
Proof: Take logs of both sides and we have,
(e/k)log(k) tends to 0 by L'Hopital's Rule as k tends to infinity. Thus, k^(e/k) tends to e^0 = 1.

Thus, lim_{k to infinity} k^(e/k)/e = 1/e < 1 implying convergence by the root test.

1) i have the series k(3/4)^k - can I say this is a converging sereis because it is a geometric sereis with abs(r) which is abs(3/4) < 1 ?

No. As it's multiplied by k, so it's not a geometric series.

Apply the root test as above. We will be left with (3/4)k^(1/k) which tends to 1 as k tends to infinity. The root test implies divergence