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    I was revising this module a while ago and wrote down a proof but now that I am going over it again, there's one step where I can't remember why I wrote what I wrote.

    The question is, Prove that if the power series  \displaystyle\sum_{n=0}^{\infty} a_nx^n converges for the value x = s then it converges for all x with   |x| < |s| .

    My answer: If  \displaystyle\sum_{n=0}^{\infty} a_nx^n converges with x = s then  a_ns^n \rightarrow 0 as  n \rightarrow \infty. Hence there is a  M \in \mathbb{R}, M \geq {0} such that  |a_ns^n| < M for all n.

    let  |z| < s and  b_n = a_nz^n.
    Then  |b_n| \leq M(\frac{|z|}{s})^n and  |\frac{z}{s}| < 1 and is constant.

    (I can't see why I wrote the last line or where it comes from)

    By ratio test, \displaystyle\sum_{n=0}^{infty} M(\frac{|z|}{s})^n converges.

    So, by comparison test, \displaystyle\sum_{n=0}^{\infty} |b_n| converges.

    And for a proof is it sufficient to just say "by so and so test" this happens or do I have to show it actually happens by applying so and so test?
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    is this a level maths? Or further maths?
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    Usually it's fine to just say "by the {insert name} test" for this kind of thing, unless you're asked specifically to prove that test.

    EDIT: To see where it comes from, apply the ratio test to the sum to the M \left( \frac{|z|}{s} \right)^n series to assure yourself it converges. Then, as your notes say, it follows by the comparison test. [If you're not sure why, read up again about the comparison test.]

    (Original post by bloomblaze)
    is this a level maths? Or further maths?
    It looks like undergraduate maths.
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    (Original post by nuodai)
    Usually it's fine to just say "by the {insert name} test" for this kind of thing, unless you're asked specifically to prove that test.

    EDIT: To see where it comes from, apply the ratio test to the sum to the M \left( \frac{|z|}{s} \right)^n series to assure yourself it converges. Then, as your notes say, it follows by the comparison test. [If you're not sure why, read up again about the comparison test.]



    It looks like undergraduate maths.
    Sorry, when I said I didn't understand why I wrote the last line, I meant the line just above the bolded part. I can't remember why I wrote and I can't see why it's true right now. Please explain it if you can, thanks. I take it that the proof is correct though?
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    (Original post by Preeka)
    Sorry, when I said I didn't understand why I wrote the last line, I meant the line just above the bolded part. I can't remember why I wrote and I can't see why it's true right now. Please explain it if you can, thanks. I take it that the proof is correct though?
    Ah right... typically I didn't read the stuff above your bolded comment :p: I think they should have said b_n = a_n \left( \frac{z}{s} \right)^n or something like that; as it stands it's definitely not right.
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    (Original post by nuodai)
    Ah right... typically I didn't read the stuff above your bolded comment :p: I think they should have said b_n = a_n \left( \frac{z}{s} \right)^n or something like that; as it stands it's definitely not right.
    I suspect it was supposed to be \left|\frac{z}{s}\right| rather than \frac{|z|}{s}. What you're aiming for is to bound |b_n| by a geometric series, so I'd expect the mod signs to be there.

    By the way, I think you slightly misinterpreted Preeka when you replied: Usually it's fine to just say "by the {insert name} test" for this kind of thing, unless you're asked specifically to prove that test.

    I totally agree with that, but the question is whether you need to actually *do* the test, or if you can just say, for example, "\sum \frac{1+n}{3+2^n} converges by the ratio test", without actually ever doing the test.

    And that's where I think you need to be careful - it depends a bit on the level you're working at. If you're doing some 3rd year analysis course, it's fine. At first year level, I would do the test (although I might write as little as "\frac{1+n+1}{3+2^{n+1}}\frac{3+2  ^n}{1+n} \to \frac{1}{2}, so sum converges by ratio test").

    However in this case:

    (Original post by Preeka)
    By ratio test, \displaystyle\sum_{n=0}^{\infty} M(\frac{|z|}{s})^n converges.
    using the ratio test is unnecessary. The expression is a geometric series with ratio < 1 and therefore convergent. (In fact, using the ratio test could be regarded as dodgy reasoning, since we often rely on the fact such a geometric converges in order to prove the ratio test).
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    (Original post by DFranklin)
    I suspect it was supposed to be \left|\frac{z}{s}\right| rather than \frac{|z|}{s}. What you're aiming for is to bound |b_n| by a geometric series, so I'd expect the mod signs to be there.

    By the way, I think you slightly misinterpreted Preeka when you replied: Usually it's fine to just say "by the {insert name} test" for this kind of thing, unless you're asked specifically to prove that test.

    I totally agree with that, but the question is whether you need to actually *do* the test, or if you can just say, for example, "\sum \frac{1+n}{3+2^n} converges by the ratio test", without actually ever doing the test.

    And that's where I think you need to be careful - it depends a bit on the level you're working at. If you're doing some 3rd year analysis course, it's fine. At first year level, I would do the test (although I might write as little as "\frac{1+n+1}{3+2^{n+1}}\frac{3+2  ^n}{1+n} \to \frac{1}{2}, so sum converges by ratio test").

    However in this case:



    using the ratio test is unnecessary. The expression is a geometric series with ratio < 1 and therefore convergent. (In fact, using the ratio test could be regarded as dodgy reasoning, since we often rely on the fact such a geometric converges in order to prove the ratio test).
    When I wrote the proof out before, I wrote it from my own revision notes so I thought I copied up wrong from somewhere. I couldn't locate it in the typed notes before but I have just found it in the typed version and it is the same as what I've written. Is the proof in the original post definitely faulty (because then I suppose I'll email the lecturer to ask) or is it just not the best most eloquent way to write the proof but still fundamentally correct?

    And also I'm still unsure where this comes from: |b_n| \leq M(\frac{z}{s})^n . (I've just quoted the original proof for the time being but I'm still not seeing why we can say |b_n| is less than the RHS.)
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    (Original post by Preeka)
    When I wrote the proof out before, I wrote it from my own revision notes so I thought I copied up wrong from somewhere. I couldn't locate it in the typed notes before but I have just found it in the typed version and it is the same as what I've written. Is the proof in the original post definitely faulty (because then I suppose I'll email the lecturer to ask) or is it just not the best most eloquent way to write the proof but still fundamentally correct?
    The exact placement of the mod signs is definitely wrong (but it's the maths equivalent of a typo - it's obvious what's meant and you probably wouldn't lost any marks for it. But there's no point learning the wrong version).

    The "by the ratio test..." bit is a little sloppy (and I think it is better to explicitly sum the GP), but again nothing to actively worry about.

    And also I'm still unsure where this comes from: |b_n| \leq M(\frac{z}{s})^n . (I've just quoted the original proof for the time being but I'm still not seeing why we can say |b_n| is less than the RHS.)
    We have b_n = a_n z^n, and |a_n s^n| &lt; M.

    Rewrite b_n as a_n s^n \frac{z^n}{s^n}, then |b_n| = |a_n s^n| \left|\frac{z}{s}\right|^n &lt; M  \left|\frac{z}{s}\right|^n
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    (Original post by DFranklin)
    The exact placement of the mod signs is definitely wrong (but it's the maths equivalent of a typo - it's obvious what's meant and you probably wouldn't lost any marks for it. But there's no point learning the wrong version).

    The "by the ratio test..." bit is a little sloppy (and I think it is better to explicitly sum the GP), but again nothing to actively worry about.

    We have b_n = a_n z^n, and |a_n s^n| &lt; M.

    Rewrite b_n as a_n s^n \frac{z^n}{s^n}, then |b_n| = |a_n s^n| \left|\frac{z}{s}\right|^n &lt; M  \left|\frac{z}{s}\right|^n
    Thank you so much
 
 
 
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