Eddygp
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Does this series converge?
Another math challenge:
How would you tell whether this series converges or does not converge (computing the sum of 10000000000 values is not allowed as a demonstration)?
[Edited on 4-4-2015 by Eddygp]
[Edited on 4-4-2015 by Eddygp]
there may be bugs in gfind
[ˌɛdidʒiˈpiː] IPA pronunciation for my Username |
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j_sum1
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well you could substitute (pi/2 - pi/2n) into the infinite series for cos.
Then you have some fun expanding and simplifying and pulling out series you know do converge. If that turns out to be the whole series then you have
your answer.
Another approach would be to apply an identity
cos(A-B) = cosAcosB + sinAsinB
then see if something interesting emerges.
I am too lazy to do either now. My guess is that it will converge since it can be shown easily that successive terms get smaller. However, (pi/2 -
pi/2n) converges to pi/2 in a manner similar to a harmonic series which just marginally diverges and cos has a point of inflection (so approaches
linear) so it could easily go the other way.
Have fun.
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j_sum1
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Forget nesrly evrrything I said above. Neither approsch will get you far.
There is a simple and elegant proof but I am not goint to yry typing it on my phone. (That's what Fermat would have said isn't it?)
It does not converge.
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Etaoin Shrdlu
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Damnit j_sum. None of mine are elegant. Come back and tell me what yours is.
Limit comparison to the harmonic series:
cos(pi/2-pi/(2n))=sin(pi/(2n)
sin(pi/(2n) is positive for n>=1
1/n is positive for n>=1
lim sin(pi/(2n))/(1/n) n->inf = pi/2
Harmonic series is divergent, so cos(pi/2-pi/(2n)) is divergent.
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woelen
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I indeed would go along the lines of Etaoin Shrdlu. For n going to infinity, the funtion can be approximated as pi/2n and the series for that is
divergent.
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gdflp
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I agree with Etaoin Shrdlu. I could explain it in layman's terms, but I'm not sure how to formally write it. | Quote: | | computing the sum of 10000000000 values is not allowed as a demonstration |
And Eddygp, I don't suppose that
has anything to do with my response to your last math problem.
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aga
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No it doesn't.
Simples !
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