Is the Alternating Series Convergent? Tips and Tricks for Solving

[SigurRos]

Alternating Series Help!

I apologize right now for the fact that I have no idea how to use LaTeX

I can't figure out if the following alternating series is convergent or not:

Sum(((-1)^(n-1)) * ((2n+1)/(n+2))) from 1 to infinity

the root test is not applicable, A(n+1)>An, and the ratio test gives me Limit=1, so I have no comclusive evidence either way. Even Maple 10 couldn't give me an answer.
I have a test tomorrow. HELP!

 

[mathphys]

You just need to use the alternating series test

the series \Sum_{n=1}^{\inftty} (-1)^n a_n converges if

lim_{n->\infty} a_n =0 & a_n is monotonic decreasing

 

[SigurRos]

Hey how can you tell if the series in monatonic decreasing?

And thank you very much.

 

[BerkMath]

That would be great if lim an=0 (n->inf) but it doesn't. lim an=2. Although the first criterion is met for the alt. series test, the second is not. In fact all tests about regular convergence fail on this series. If you are allowed to broaden your idea of convergence it may be Abel or Cesaro summable. Otherwise, every other test for "regular" convergence is inconclusive. Unless there is a trick I am missing you might not know whether it converges in traditional sense.

 

[BerkMath]

lol, it just killed Matlab 7.0 too! "NaN"

 

[mathphys]

If {a_n} is a (strictly) monotonically decreasing series then

(a_(n+1))/a_n < 1 . (for all n)

For the alternating test both this and the limit must hold,
hence if one fails, the series does not converge

 

[SigurRos]

An+1>An
and
the Lim=2 as n->infinity

Does this mean that it is definitely divergent then, since neither criteria for convergence was met?

 

[BerkMath]

mathphys, are we going to argue all night. I will state the alternating series test: An alternating series sum((-1)^n*an) converges IF: an is monotone decreasing AND lim(an)=0. We have the prepositional logic statement: (A+B)-->C. Now you are saying that if we negate either A or B, ~A or ~B, then that means ~C? WLOG, say we have A and ~B. Then we have ~(A+B). So your claim is that given (A+B)->C and ~(A+B), then we may conclude ~C. That is a logical fallacy called denying the antecedent. Therefore it is still unknown whether the series converges or diverges.

 

[SigurRos]

Hate to be a burden here, but I have 2 Power series questions as well

I have to find the radius of convergence for each of these:

((sqrt(n)+1)/(2+sqrt(n))*x^n

and

((2n)!/((n+1)^n))*x^n

Im sure I could work them out, but the test is in 4 hours and my brain ceases functioning at this hour.

Thanks for your help so far, guys.

 

[mathphys]

"For the alternating test both this and the limit must hold,
hence if one fails, the series does not converge", i wrote, "hence if that one...", I must had written ;)... meaning that if the limit criteria is not satisfied the series diverge.

Lim_{n-->infinity} a_n=0 is a necessary condition for convergence of the alternating series. If this criteria fails the series diverge. Nevertheless, if this requierement is satisfied this does not mean that the series converge, because is necessary but not sufficient.

However, in this example this limit does not hold, so the series does not converge (for THIS example).

 

[BerkMath]

For the first one: B=lim(sup(abs(an)^(1/n),N),N,inf)=1. Therefore 1/B=1=Radius of convergence. For the other the test fails, so give me a moment.

 

[BerkMath]

Using the ratio test on the second one, lim(abs(a(n)/a(n+1))=1/2. Therefore R=1/2.

 

[BerkMath]

mathphys, you fail to understand the logic behind conditional statements. The only inferences that may be made using the conditional statement A->B, are Modus Pollens and Modus Tollens. In the former, we have A->B, A, therefore B. In the latter A->B, ~B, therefore ~A. We have neither of the two cases in this situation and by continuing to support the claim that since lim an=2 then the series diverges, shows that you are unwilling to put as much effort in understanding prepositional logic as you are to argue with people on the internet. You are commiting a logical fallacy, denying the antecedent, by saying for certain it converges. Example: "If I was killed in a plane crash, then I am dead." I was not killed in a plane crash. By your logic, we may infer that I am not dead? Does that make any sense?

 

[mathphys]

out

I won't discuss logic with you, because this is not the place or post, nor i will put simplistic non sense examples.


Just try understand what necessary and sufficient conditions mean.

 

[BerkMath]

My friend, you are wrong. I gave a proof of why you are wrong. (A+B)->C, ~(A+B), therefore ~C is INCORRECT, denying the antecedent. Perhaps I didn't make clear what A, B, and C are. A=an is monotonically decreasing, B=lim(an)=0, and C=an converges. Now we know A to be true, we know B to be false, therefore ~B. Now, in order for A+B to be true, BOTH A and B must be true, therefore we have ~(A+B). It is therefore INCONCLUSIVE whether we have C or ~C. Give me a PROOF otherwise? You claim that I don't know necessary and sufficient conditions, give me a proof of why I am wrong. And by the way, my example wasn't stupid. just because I didn't get killed in a plane crash, doesn't mean that I am not dead, I could have died by other means. Perhaps going insane on the internet why attmepting to persuade the thick headed.

 

[SigurRos]

...i didnt mean to start a fight
i just wanted homework help
feel free to continue though, I am not really knowledgeable on logic and proofs, so maybe i can learn something from this debate

Thanks again for your help guys

 

[mathphys]

Lol

There is no point in that , it won't help you with your homework, nor me. :)
Just for reference
http://planetmath.org/encyclopedia/AlternatingSeriesTest.html see this and references therein.

http://en.wikipedia.org/wiki/Necessary_and_sufficient_conditions

"To say that A is necessary for B is to say that B cannot be true unless A is true, or that whenever (wherever, etc.) B is true, so is A."

Because the limit condition ALONE (not A+B nor nothing like that :P) is a necessary condition for the alternating series to converge, it cannot be true that the series converge unless the limit condition (ALONE!) is satified. Because it is not, is not true that the series converge. Then what follows? you guess it :).

 

[SigurRos]

wait, for the second power series, after some manipulation, wouldn't you end up with 2x* Lim((1+1/n)/(1+2/n))^(n+1), which would be 2x/e, and the radius would be e/2?

 

[mathphys]

:)

You're on the right track, an exponential has to appear, just work it a little more

 

[SigurRos]

well (1+1/n)^(n+1) is e
and 1 + 2/n)^(n+1) is e^2
so it would have to be 2x*e^-1, or 2x/e, right?
or am i missing something
2 hours till test...
i love caffeine

 

[mathphys]

hint

Check the definition of the exponential of a number as a limit.

Also take care with the factorial

(2n)! turns in (2(n+1))! when n is replaced by n+1

 

[SigurRos]

im pretty sure the factorial is right because when (2(n+1))! is divided by (2n)!, it equals (2n)!*2(n+1)/(2n)!, which cancells out to 2(n+1)
...right?
and i checked that limit with maple and it gave me e^-1

wait a minute, is 4x/e, therefore x=e/4 because of the manner of this particular factorial?

 

[mathphys]

factorial

I think that
(2(n+1))!= (2n+2)!=2n! (2n+1)(2n+2)

unless that I may need some caffeine also. (I think I do

 

[SigurRos]

i don't think that's so, because all n is replaced with (n+1), making (2n)->2(n+1)->2n+2. To get 2n+1 you'd have to add a n value of 1/2.
if you divide out 2n+1, you'd get 2(n+1/2) which is incorrect, i think.
im pretty sure that the way this series works, it is only 2n+2, or 2(n+1)

i could be wrong, and ill make sure to let you know who's right when i get the test back.
maybe the answer is just 1,000,000,000*pi and we both seriously need caffeine.
only time can tell.

 

[mathphys]

;)

It is correct, believe me, after all the factorial is just a particular case of the gamma function ;)

For example

(2(3)+2)!= 8!=8*7*(6*5*4*3*2*1*1)= 8*7*6!= (2*3+2)(2*3+1)*6!



And Yes,,,long live to caffeine!

 

[SigurRos]

i hereby retract my previous statement

you are right, mathphys
i was wrong

thank you for showing me the way
(calculators don't lie)

 

[SigurRos]

with this new knowledge, imust now include (2n+1)in my limit, which is not pretty, as it produces infinity.
correct?

 

[SigurRos]

and therefore, i end up with 2x/e * infinity, and therefore the radius of convergence is 0
right?

 

[mathphys]

Yes

It converges just at x=0

 

[SigurRos]

awesome!

you rule
as do i, but to a clearly lesser extent

 

[mathphys]

;)

Well, We all have our 'I need help' problems, it is just a matter of practice and becoming every day a little bit more masochist and stubborn in order to solve them LOL (I'm stuck on one of those now). And because of that, it is nice to have the help and hints when are needed.

Good Luck, in your examination!