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    having a bit of bother proving the following:

    For each positive integer n, let a_n be a real number with a_n[-11,23] and define the sequenceb (x_n)_n by

    \displaystyle x_n=\frac{a_1cos(e)}{1^2}+\frac{  a_2cos(e^2)}{2^2}+...+\frac{a_nc  os(e^n)}{n^2}

    show the sequence (x_n)_n is cauchy.

    only a 4 mark question on a paper of ~100 marks so it cant be that intensive a question, I think I'm just missing something.... If anyone could point out what to do I would appreciate it.
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    (Original post by Square)
    having a bit of bother proving the following:

    For each positive integer n, let a_n be a real number with a_n[-11,23] and define the sequenceb (x_n)_n by

    \displaystyle x_n=\frac{a_1cos(e)}{1^2}+\frac{  a_2cos(e^2)}{2^2}+...+\frac{a_nc  os(e^n)}{n^2}

    show the sequence (x_n)_n is cauchy.

    only a 4 mark question on a paper of ~100 marks so it cant be that intensive a question, I think I'm just missing something.... If anyone could point out what to do I would appreciate it.
    |\frac{a_ncos(e^n)}{n^2}|<|\frac  {a_n}{n^2}|

    See anything?
    Edit-forgot to add, it should be a less than or equals sign, but I don't know how to do that...
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    If a_n is a positive Cauchy sequence and |b_n| \leq a_n \forall n then b_n is Cauchy. (You should be able to prove this easily).
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    (Original post by Slumpy)
    |\frac{a_ncos(e^n)}{n^2}|<|\frac  {a_n}{n^2}|

    See anything?
    Edit-forgot to add, it should be a less than or equals sign, but I don't know how to do that...
    no I got this, I'm just getting confused with all the a_n's what do I do with them?
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    (Original post by Square)
    no I got this, I'm just getting confused with all the a_n's what do I do with them?
    Look at DFranklin's hint above.
    You're trying to show that the x_n get closer and closer together, so a way to show that would be to show that the terms being added on are getting sufficiently smaller.
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    i still dont follow, im pretty bad at this!

    Here is my working so far.

    epsilon>0, N natural number s.t n,m>N |x_n-x_m|\leq \epsilon

    \displaystyle |\frac{a_1cose}{1^2} + ... \frac{a_ncose^n}{n^2}-(\frac{a_1cose}{1^2} + ... \frac{a_mcose^m}{m^2})|

    =\displaystyle |\frac{a_{m+1}cose^{m+1}}{(m+1)^  2} + ... \frac{a_ncose^n}{n^2}|

    \leq \displaystyle |\frac{a_{m+1}}{(m+1)^2} + ... \frac{a_n}{n^2}|
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    (Original post by Slumpy)
    Look at DFranklin's hint above.
    You're trying to show that the x_n get closer and closer together, so a way to show that would be to show that the terms being added on are getting sufficiently smaller.
    quoting so you see this hopefully, check my post above!
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    You know \sum_1^\infty \frac{1}{n^2} converges, so it's Cauchy. Now compare a multiple of this with your series.
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    (Original post by Square)
    i still dont follow, im pretty bad at this!

    Here is my working so far.

    epsilon>0, N natural number s.t n,m>N |x_n-x_m|\leq \epsilon

    \displaystyle |\frac{a_1cose}{1^2} + ... \frac{a_ncose^n}{n^2}-(\frac{a_1cose}{1^2} + ... \frac{a_mcose^m}{m^2})|

    =\displaystyle |\frac{a_{m+1}cose^{m+1}}{(m+1)^  2} + ... \frac{a_ncose^n}{n^2}|

    \leq \displaystyle |\frac{a_{m+1}}{(m+1)^2} + ... \frac{a_n}{n^2}|
    That's you basically there really.
    If you prove the result Dave suggested though, you could consider the sequence x_n=a|frac{1}{1^2}+...+frac{1}{n  ^2} | where a is the largest value a_n can take.
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    I guess this is where the interval on an comes in? So would I want to consider 24a_n (a_n=sum(1/n^2)), given that would be larger than b_n (the sequence above) for all n?
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    (Original post by Square)
    I guess this is where the interval on an comes in? So would I want to consider 24a_n (a_n=sum(1/n^2)), given that would be larger than b_n (the sequence above) for all n?
    That would work, yes.
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    (Original post by Slumpy)
    That would work, yes.
    well i really made a meal of that, thanks for the help guys! rep headed your way.
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    (Original post by Square)
    well i really made a meal of that, thanks for the help guys! rep headed your way.
    No problem, helped me avoid my own work for at least a couple of minutes:p:
 
 
 
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