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    Doing these questions and would appreciate for someone to check my answers and/or guide me through the one's I don't understand.

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    B1: (a_n)\rightarrow \infty if, and only if, \forall C>0, \exists N \in \mathbb{N} s.t. \forall n> N, a_n>C


    B2: Here I am unsure. The only thing about this in my notes is the d'Alembert's ratio test but that seems to explain the behaviour of (a_n) as l differs under 5 different scenarios. It doesn't look like a "definition" to me. :/

    The only thing I can think of is to go the other way such as "If (a_n)\rightarrow \infty then it implies that l>1 for \frac{a_{n+1}}{a_n}\rightarrow l" assuming the converse of these statements is true.


    B3: (a_n) \rightarrow \infty and (b_n)\rightarrow 0


    B4: Part 5 of d'Alembert's ratio tests says "If l=1 we get no information"

    (\frac{a_{n+1}}{a_n})=(\frac{2(n  +1)+3}{2n+3})=(1+\frac{2}{2n+3})  =(1+2b_n) and since (b_n) \rightarrow 0 \Rightarrow (2b_n)\rightarrow 0 by the constant multiple rule. This concludes that (\frac{a_{n+1}}{a_n})\rightarrow 1 and we get no further information about the sequence's divergence/convergence??? Which is strange because it should l should be greater than 1 for (a_n) to tend to infinity. So I'm confused.

    Similarly for \displaystyle \frac{b_{n+1}}{b_n}=(\frac{2n+3}  {2n+5})=(\frac{2+\frac{3}{n}}{2+  \frac{5}{n}}) so this implies that (\frac{b_{n+1}}{b_n})\rightarrow 1 which again I am confused about as there's a contradiction.

    Thanks in advance.

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    (Original post by RDKGames)
    Doing these questions and would appreciate for someone to check my answers and/or guide me through the one's I don't understand.

    Name:  asdf.PNG
Views: 72
Size:  16.6 KB

    B1: (a_n)\rightarrow \infty if, and only if, \forall C>0, \exists N \in \mathbb{N} s.t. \forall n> N, a_n>C


    B2: Here I am unsure. The only thing about this in my notes is the d'Alembert's ratio test but that seems to explain the behaviour of (a_n) as l differs under 5 different scenarios. It doesn't look like a "definition" to me. :/

    The only thing I can think of is to go the other way such as "If (a_n)\rightarrow \infty then it implies that l>1 for \frac{a_{n+1}}{a_n}\rightarrow l" assuming the converse of these statements is true.


    B3: (a_n) \rightarrow \infty and (b_n)\rightarrow 0


    B4: Part 5 of d'Alembert's ratio tests says "If l=1 we get no information"

    (\frac{a_{n+1}}{a_n})=(\frac{2(n  +1)+3}{2n+3})=(1+\frac{2}{2n+3})  =(1+2b_n) and since (b_n) \rightarrow 0 \Rightarrow (2b_n)\rightarrow 0 by the constant multiple rule. This concludes that (\frac{a_{n+1}}{a_n})\rightarrow 1 and we get no further information about the sequence's divergence/convergence??? Which is strange because it should l should be greater than 1 for (a_n) to tend to infinity. So I'm confused.

    Similarly for \displaystyle \frac{b_{n+1}}{b_n}=(\frac{2n+3}  {2n+5})=(\frac{2+\frac{3}{n}}{2+  \frac{5}{n}}) so this implies that (\frac{b_{n+1}}{b_n})\rightarrow 1 which again I am confused about as there's a contradiction.

    Thanks in advance.

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    B3 is right, and B4 looks okay to me too. I'm not sure about the definitions, you should have those in your course notes.

    For the ratio test for an, yeah it should be >1 for it tending to infinity, but as it =1 we don't know what the sequence does.

    I think this question is just trying to show you examples of where the ratio test fails.
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    (Original post by RDKGames)
    B2: Here I am unsure. The only thing about this in my notes is the d'Alembert's ratio test but that seems to explain the behaviour of (a_n) as l differs under 5 different scenarios. It doesn't look like a "definition" to me. :/

    The only thing I can think of is to go the other way such as "If (a_n)\rightarrow \infty then it implies that l>1 for \frac{a_{n+1}}{a_n}\rightarrow l" assuming the converse of these statements is true.
    This is just a standrd bookwork "definition of a limit", it might make it more concrete if you write b_n = \frac{a_{n+1}}{a_n} then you want to write down what it means for (b_n) \to \ell. In particular, it just means that for all positive real \epsilon there exists a natural N such that whenever n > N we have |b_n - \ell| < \epsilon.

    You can then 'back-substitute' for a_n.

    On a stylistic note, by the way - it's much more lucid and readable if you write your definitions in terms of actual words instead of overusing the notation definition - I know you've just learnt it and that sort of unconsciously makes you want to use it more, but you'll find that it's much more readable when done in mainly words.
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    (Original post by RDKGames)
    B4: Part 5 of d'Alembert's ratio tests says "If l=1 we get no information"

    (\frac{a_{n+1}}{a_n})=(\frac{2(n  +1)+3}{2n+3})=(1+\frac{2}{2n+3})  =(1+2b_n) and since (b_n) \rightarrow 0 \Rightarrow (2b_n)\rightarrow 0 by the constant multiple rule. This concludes that (\frac{a_{n+1}}{a_n})\rightarrow 1 and we get no further information about the sequence's divergence/convergence??? Which is strange because it should l should be greater than 1 for (a_n) to tend to infinity. So I'm confused.

    Similarly for \displaystyle \frac{b_{n+1}}{b_n}=(\frac{2n+3}  {2n+5})=(\frac{2+\frac{3}{n}}{2+  \frac{5}{n}}) so this implies that (\frac{b_{n+1}}{b_n})\rightarrow 1 which again I am confused about as there's a contradiction.
    For this part, I'm doing it in a separate post to make it more lucid, there's no contradiction. You're misunderstanding the ratio test.

    The ratio test says that: IF \ell > 1 THEN it diverges. It doesn't say IF it diverges THEN \ell > 1. Those are two seperate statements that mean two very different things. So a sequence could still diverge but have \ell \ngtr 1 (in particular \ell =1).

    In particular, the 'weak point' of the ratio test is that \ell = 1 means absolutely nothing, the sequence could diverge, the sequence could converge, you just don't know - the ratio test essentially spatters out "use another test, I'm not useful here.".
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    (Original post by Zacken)
    This is just a standrd bookwork "definition of a limit", it might make it more concrete if you write b_n = \frac{a_{n+1}}{a_n} then you want to write down what it means for (b_n) \to \ell. In particular, it just means that for all positive real \epsilon there exists a natural N such that whenever n > N we have |b_n - \ell| < \epsilon.

    You can then 'back-substitute' for a_n.
    Ah yes, that form with the help of substitution makes it much clearer and I found the notes on that.

    Just for verification, so if I was to rewrite the definition for it, would this be the correct way? And I assume I could always just skip the substitution for the sake of the final definition? I think I'd write this full of symbols but I'll keep it in words here, as it's very similar to what you've said.

    Let (b_n)=\frac{a_{n+1}}{a_n} then it follows that (b_n)\rightarrow \ell if, and only if, for all \epsilon > 0 there exists N in the naturals, such that for all n>N we have \lvert b_n-\ell \lvert < \epsilon

    On a stylistic note, by the way - it's much more lucid and readable if you write your definitions in terms of actual words instead of overusing the notation definition - I know you've just learnt it and that sort of unconsciously makes you want to use it more, but you'll find that it's much more readable when done in mainly words.
    I see. I've gotten too used to using these symbols and while the output is much more concise, it may not always be clear. I'll try to tone down on it. :lol:

    Our lecturer wanted us to get used to all the notation in the first few weeks so what better way than to throw it around :ninja:

    For this part, I'm doing it in a separate post to make it more lucid, there's no contradiction. You're misunderstanding the ratio test.

    The ratio test says that: IF THEN it diverges. It doesn't say IF it diverges THEN . Those are two seperate statements that mean two very different things. So a sequence could still diverge but have (in particular ).

    In particular, the 'weak point' of the ratio test is that means absolutely nothing, the sequence could diverge, the sequence could converge, you just don't know - the ratio test essentially spatters out "use another test, I'm not useful here.".
    Ah right - yeah I misunderstood it slightly. So just as said by rayquaza17, this part is most likely showcasing an example of when this test fails and I assume this is what we're expected to pick up upon.


    Thanks for the help!
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    (Original post by RDKGames)
    Ah yes, that form with the help of substitution makes it much clearer and I found the notes on that.

    Just for verification, so if I was to rewrite the definition for it, would this be the correct way? And I assume I could always just skip the substitution for the sake of the final definition? I think I'd write this full of symbols but I'll keep it in words here, as it's very similar to what you've said.

    Let (b_n)=\frac{a_{n+1}}{a_n} then it follows that (b_n)\rightarrow \ell if, and only if, for all \epsilon > 0 there exists N in the naturals, such that for all n>N we have \lvert b_n-\ell \lvert < \epsilon
    Yep, that's a good way of saying it - not quite the final definition though. You want the inequality to be \left|\frac{a_{n+1}}{a_n} - \ell \right| < \epsilon which you can simplify a little.

    Ah right - yeah I misunderstood it slightly. So just as said by rayquaza17, this part is most likely showcasing an example of when this test fails and I assume this is what we're expected to pick up upon.
    I presume so, yes - I guess it's also an exposition of the essential difference between a theorem and it's converse.
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    (Original post by Zacken)

    In particular, the 'weak point' of the ratio test is that \ell = 1 means absolutely nothing, the sequence could diverge, the sequence could converge, you just don't know - the ratio test essentially spatters out "use another test, I'm not useful here.".
    Just to amplify this, the "point" of them asking you about the ratio limit of those sequences is to emphasise that you *really* get no information when l=1. That is, it's not just that the theorem gives no information (which leaves open the possibility that some 'cleverer' theorem might do better). It's that there are both divergent and convergent series with l=1.
 
 
 
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