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    Hi,

    I'm attempting to solve this problem

    http://imgur.com/yjpxUgp

    I have tried to solve it but can't prove the result: http://imgur.com/Kk2VplG

    Any help would be much appreciated

    Many thanks
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    (Original post by Radio10)
    Hi,

    I'm attempting to solve this problem

    http://imgur.com/yjpxUgp

    I have tried to solve it but can't prove the result: http://imgur.com/Kk2VplG

    Any help would be much appreciated

    Many thanks
    Hint: From \displaystyle S_N = \sum_{n=1}^N \dfrac{n}{2^{n-1}}, write k = n-1 to get

    \displaystyle S_N = \sum_{k=0}^{N-1} \dfrac{k+1}{2^k}

    Split this into 2 sums (one with k/2^k and one with 1/2^k). You can sum the latter, and manipulate the former to look like the summation you are trying to get to.
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    (Original post by DFranklin)
    Hint: From \displaystyle S_N = \sum_{n=1}^N \dfrac{n}{2^{n-1}}, write k = n-1 to get

    \displaystyle S_N = \sum_{k=0}^{N-1} \dfrac{k+1}{2^k}

    Split this into 2 sums (one with k/2^k and one with 1/2^k). You can sum the latter, and manipulate the former to look like the summation you are trying to get to.
    Thank you very much for your quick response.

    I have managed to solve the latter part with your help: http://imgur.com/a/PdQAf

    For the former part, do I say when k=0, n=1 and change it to look like the result given?

    Many thanks for your help
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    (Original post by Radio10)
    Thank you very much for your quick response.

    I have managed to solve the latter part with your help: http://imgur.com/a/PdQAf

    For the former part, do I say when k=0, n=1 and change it to look like the result given?
    You can "relabel" a sum (that is, change the variable you're summing over) without changing it.

    So \sum_{k=1}^N f(k) is exactly the same thing as \sum_{n=1}^N f(n).

    So relabel k to n, you then have something very like what you want - the only issue is the sum goes from 0 instead of 1. But what you're summing is 0 when n (or k) = 0, so you can simply remove that term.
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    (Original post by DFranklin)
    You can "relabel" a sum (that is, change the variable you're summing over) without changing it.

    So \sum_{k=1}^N f(k) is exactly the same thing as \sum_{n=1}^N f(n).

    So relabel k to n, you then have something very like what you want - the only issue is the sum goes from 0 instead of 1. But what you're summing is 0 when n (or k) = 0, so you can simply remove that term.
    Ah I see. Thank you for that explanation.

    I apologise for posting my previous workings as attachments. I hadn't thought about the hassle of switching between tabs when responding to posts. Sorry about that. Will try to post in plain text/LaTeX next time.

    If I'm not troubling you could you give a hint on how to solve part b. I'm not quite sure on it.
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    (Original post by Radio10)
    I apologise for posting my previous workings as attachments. I hadn't thought about the hassle of switching between tabs when responding to posts. Sorry about that. Will try to post in plain text/LaTeX next time.
    Please don't take it personally - despite the tone of the thread, we all recognize that many people don't know LaTeX. Although I think you probably should learn it (over time) if you're doing a maths degree, as long as you are doing the other things that make your posts worth replying to (like making effort of your own, posting working, etc., all of which you are doing), I'm not going to have a problem with it.

    If I'm not troubling you could you give a hint on how to solve part b. I'm not quite sure on it.
    All they've done is divide S_N by 2 and separate out the last term (when n = N) from the sum.
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    (Original post by DFranklin)
    Please don't take it personally - despite the tone of the thread, we all recognize that many people don't know LaTeX. Although I think you probably should learn it (over time) if you're doing a maths degree, as long as you are doing the other things that make your posts worth replying to (like making effort of your own, posting working, etc., all of which you are doing), I'm not going to have a problem with it.

    All they've done is divide S_N by 2 and separate out the last term (when n = N) from the sum.
    It's fine, I understand the point you make

    Oh I see, I was overcomplicating it

    Please could you have a look at my attempt for part c. I can't seem to find where I am going wrong.
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    (Original post by Radio10)
    It's fine, I understand the point you make

    Oh I see, I was overcomplicating it

    Please could you have a look at my attempt for part c. I can't seem to find where I am going wrong.
    Sorry, but I really can't work out what you're trying to do here. You've written a bunch of stuff, but nothing seems to connect one line to anoher, or even indicate what you're actually trying to evaluate. (I think this is something A-level doesn't teach well, but you need to do it at university level. It's a bit like doing English comprehension - once you get to a certain level, you need to start writing in full (mathematical) sentences),

    I can tell you that at this point, you shouldn't be doing any more manipulation of the sums at all.

    If we write A for S_n, and B for \sum_1^{n-1} \frac{n}{2^n}, then (a) and (b) can be rewritten as:

    (a) A = B + C, (where C = 2(1- 1/2^N))
    (b) B = A/2 + D (where D = N / 2^N).

    Treat these as simultaneous equations in A and B, and solve to find A in terms of C and D.
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    (Original post by DFranklin)
    Sorry, but I really can't work out what you're trying to do here. You've written a bunch of stuff, but nothing seems to connect one line to anoher, or even indicate what you're actually trying to evaluate. (I think this is something A-level doesn't teach well, but you need to do it at university level. It's a bit like doing English comprehension - once you get to a certain level, you need to start writing in full (mathematical) sentences),

    I can tell you that at this point, you shouldn't be doing any more manipulation of the sums at all.

    If we write A for S_n, and B for \sum_1^{n-1} \frac{n}{2^n}, then (a) and (b) can be rewritten as:

    (a) A = B + C, (where C = 2(1- 1/2^N))
    (b) B = A/2 + D (where D = N / 2^N).

    Treat these as simultaneous equations in A and B, and solve to find A in terms of C and D.
    I really appreciate your quick response. I'm sorry that my previous post didn't make sense, I was trying to substitute the RHS of part b into the RHS of part a and simplify but it seems that wouldn't have been useful.

    Thank you so much for the above. I was able to finally prove the result. I will type it up in a bit.

    I have one last question (part d) please
    S_N= 4 - (N+2)/ 2^(N-1)

    Would it be sufficient to write
    (N+2)/ 2^(N-1)= N/2^ (N-1) + 2/2^(N-1)

    =2N/2^N + 4/2^N

    For the latter part, as N tends to infinity, 4/2^N tends to zero.

    For the former, this converges to 0, because as n gets bigger, the function will get closer and closer to 0
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    (Original post by Radio10)
    I really appreciate your quick response. I'm sorry that my previous post didn't make sense, I was trying to substitute the RHS of part b into the RHS of part a and simplify but it seems that wouldn't have been useful.
    I'm not sure how much it would have helped here (because I think some other things went wrong at the same time!), but in general the kind of explanation you've just given makes things *so* much easier for your helpers to follow...

    Would it be sufficient to write
    (N+2)/ 2^(N-1)= N/2^ (N-1) + 2/2^(N-1)

    =2N/2^N + 4/2^N

    For the latter part, as N tends to infinity, 4/2^N tends to zero.

    For the former, this converges to 0, because as n gets bigger, the function will get closer and closer to 0
    This isn't sufficient. In particular your treatment of "the former part" is insufficient.

    The question explicitly tells you that "you will need to say something about the convergence of \frac{N}{2^N} as N\to \infty". Moreover, it gives you a hint (and so you can't simply say "I considered it and it tends to 0".

    [Also, I'll note that 1+(1/x) "gets closer and closer to 0" as x goes to infinity, but it sure as heck doesn't tend to 0...]

    On the other hand, I can't say I find their hint terribly useful.

    If I wanted to prove N/2^N \to 0 I would go (there are details in the following you'll need to fill and, as well as all the bits where I've just written "..." rather than give the game away):

    Let a_n = n / 2^n. Then \frac{a_{n+1}}{a_n} = ... So if n > ... then \frac{a_{n+1}}{a_n} < \frac{3}{4}. Then by comparison with the geometric series AR^n (with A = ..., R = ...), a_n tends to 0.

    As I say, I really don't get the hint. (I can think of ways to use it, but they feel inappropriate given the rest of the Q).
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    (Original post by DFranklin)
    I'm not sure how much it would have helped here (because I think some other things went wrong at the same time!), but in general the kind of explanation you've just given makes things *so* much easier for your helpers to follow...

    This isn't sufficient. In particular your treatment of "the former part" is insufficient.

    The question explicitly tells you that "you will need to say something about the convergence of \frac{N}{2^N} as N\to \infty". Moreover, it gives you a hint (and so you can't simply say "I considered it and it tends to 0".

    [Also, I'll note that 1+(1/x) "gets closer and closer to 0" as x goes to infinity, but it sure as heck doesn't tend to 0...]

    On the other hand, I can't say I find their hint terribly useful.

    If I wanted to prove N/2^N \to 0 I would go (there are details in the following you'll need to fill and, as well as all the bits where I've just written "..." rather than give the game away):

    Let a_n = n / 2^n. Then \frac{a_{n+1}}{a_n} = ... So if n > ... then \frac{a_{n+1}}{a_n} < \frac{3}{4}. Then by comparison with the geometric series AR^n (with A = ..., R = ...), a_n tends to 0.

    As I say, I really don't get the hint. (I can think of ways to use it, but they feel inappropriate given the rest of the Q).
    I'll be sure to add an explanation alongside my working in future. Thank you for the advice.



    Please let me know if the following is correct or which areas I need to amend.


    Let a_n = n / 2^n. Then \frac{a_{n+1}}{a_n} = \frac{n+2^{n}}{n}. So if n > 0 then \frac{a_{n+1}}{a_n} < \frac{3}{4}. Then by comparison with the geometric series AR^n (with A = 1, R = 3/4), a_n tends to 0.

    I feel I have made a mistake (which I probably have).
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    (Original post by Radio10)
    ..
    Your ratio isn't correct.

    There's a LaTeX "issue" in what I wrote that I think may be confusing you (the LaTeX I wrote is "correct", but the LaTeX output isn't as clear as I'd hoped).

    When I wrote \frac{a_{n+1}}{a_n}, I meant \dfrac{a_{n+1}}{a_n} (i.e. the ratio of the nth and n+1th terms of a_n). I think you may have interpreted it as \dfrac{a_n+1}{a_n}.

    Anyhow, a_{n+1} has a factor (n+1), while a_n has a factor n. So you'd better have something looking like (n+1)/n as part of your ratio. .

    Also, I'm not sure where your A came from, or if it's right. To give a slightly different example (so I'm not giving you a full solution): suppose we'd shown that

    a_{n+1} < (3/4) a_n for n = 7.

    I would write something like: since a_{n+1} < (3/4) a_n for n = 7, we can deduce a_n &gt; (3/4)^{n-7} a_7 for n = 7. So by comparison with the GP AR^n with A = a_7 (4/3)^7, we can deduce...

    To be fair, you don't really need to know what A is, as long as |R| < 1, you're fine.
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    (Original post by DFranklin)
    Your ratio isn't correct.

    There's a LaTeX "issue" in what I wrote that I think may be confusing you (the LaTeX I wrote is "correct", but the LaTeX output isn't as clear as I'd hoped).

    When I wrote \frac{a_{n+1}}{a_n}, I meant \dfrac{a_{n+1}}{a_n} (i.e. the ratio of the nth and n+1th terms of a_n). I think you may have interpreted it as \dfrac{a_n+1}{a_n}.

    Anyhow, a_{n+1} has a factor (n+1), while a_n has a factor n. So you'd better have something looking like (n+1)/n as part of your ratio. .

    Also, I'm not sure where your A came from, or if it's right. To give a slightly different example (so I'm not giving you a full solution): suppose we'd shown that

    a_{n+1} < (3/4) a_n for n = 7.

    I would write something like: since a_{n+1} < (3/4) a_n for n = 7, we can deduce a_n &gt; (3/4)^{n-7} a_7 for n = 7. So by comparison with the GP AR^n with A = a_7 (4/3)^7, we can deduce...

    To be fair, you don't really need to know what A is, as long as |R| < 1, you're fine.
    Sorry I did misinterpret it. Is the correct ratio n+1/2n with n>2?

    Oh ok. I will work on finding the correct value of A
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    (Original post by Radio10)
    Sorry I did misinterpret it. Is the correct ratio n+1/2n with n>2?
    Assuming you mean \frac{n+1}{2n} and not n+\frac{1}{2n} then yes!

    Oh ok. I will work on finding the correct value of A
    As I say, the exact value doesn't really matter, but you should probably do it once correctly so you know you're not doing anything wrong.
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    (Original post by DFranklin)
    Assuming you mean \frac{n+1}{2n} and not n+\frac{1}{2n} then yes!

    As I say, the exact value doesn't really matter, but you should probably do it once correctly so you know you're not doing anything wrong.
    Yes, I will make sure I start to learn LaTeX soon.

    Will do. Thank you so much for your help throughout this question. I really appreciate it!!!!

    Many thanks
 
 
 
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