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    Hello,

    Please could you check if my method is correct?


    A sequence is defined by the recurrence relation u_{n+1}=-\frac{5}{u_n}+6 with u1 = 2

    Prove by induction that for any natural number n, 1<u_n<5


    Step 1: Check for n=1. We know that u1=2, so 1<u1<5

    Step 2: Assume that the inequality holds for all natural n.

    Step 3: Check for n+1.
    We have
    u_{n+1}=-\frac{5}{u_n}+6

    If we sub. un for 1, we get
    u_{n+1}=-\frac{5}{1}+6 =1

    If we sub. un for 5, we get
    u_{n+1}=-\frac{5}{5}+6 =5

    Since we had assumed that 1<un<5, we conclude that 1<Un+1<5.

    We have shown that the inequality holds for n+1 and thus it holds for all natural n.


    Any clarification is much appreciated.
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    (Original post by razzor)
    Hello,

    Please could you check if my method is correct?


    A sequence is defined by the recurrence relation u_{n+1}=-\frac{5}{u_n}+6 with u1 = 2

    Prove by induction that for any natural number n, 1&lt;u_n&lt;5


    Step 1: Check for n=1. We know that u1=2, so 1<u1<5

    Step 2: Assume that the inequality holds for all natural n.

    Step 3: Check for n+1.
    We have
    u_{n+1}=-\frac{5}{u_n}+6

    If we sub. un for 1, we get
    u_{n+1}=-\frac{5}{1}+6 =1

    If we sub. un for 5, we get
    u_{n+1}=-\frac{5}{5}+6 =5

    Since we had assumed that 1<un<5, we conclude that 1<Un+1<5.

    We have shown that the inequality holds for n+1 and thus it holds for all natural n.


    Any clarification is much appreciated.
    THat is correct but I should check for n=2 because n=1 -> u1 is only a given initial value. We can get for n=2 ->u2 a calculated value through the recurrent relation.
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    Thanks a lot for your speedy reply
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    (Original post by razzor)
    Hello,

    Please could you check if my method is correct?


    A sequence is defined by the recurrence relation u_{n+1}=-\frac{5}{u_n}+6 with u1 = 2

    Prove by induction that for any natural number n, 1&lt;u_n&lt;5


    Step 1: Check for n=1. We know that u1=2, so 1<u1<5

    Step 2: Assume that the inequality holds for all natural n.

    Step 3: Check for n+1.
    We have
    u_{n+1}=-\frac{5}{u_n}+6

    If we sub. un for 1, we get
    u_{n+1}=-\frac{5}{1}+6 =1

    If we sub. un for 5, we get
    u_{n+1}=-\frac{5}{5}+6 =5

    Since we had assumed that 1<un<5, we conclude that 1<Un+1<5.

    We have shown that the inequality holds for n+1 and thus it holds for all natural n.


    Any clarification is much appreciated.
    Unfortunately you haven't proved it at all.

    You've shown that 1 and 5 map to 1 and 5, but you haven't shown that for some value between 1 and 5 you couldn't output a value that goes outside that range, which is what the question is asking you to do!

    Also: just noticed for Step 2 you've said "assume inequality holds for all natural numbers n". That is NOT how you set up the induction hypothesis! You should be saying "assume the inequality holds for a given natural number k". Your task is then to show that it holds for k+1 (that's what you've actually tried to do, but your wording is incorrect and you shouldn't use 'n' because 'n' refers to the general term).
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    (Original post by razzor)
    Hello,

    Please could you check if my method is correct?


    A sequence is defined by the recurrence relation u_{n+1}=-\frac{5}{u_n}+6 with u1 = 2

    Prove by induction that for any natural number n, 1&lt;u_n&lt;5


    Step 1: Check for n=1. We know that u1=2, so 1<u1<5

    Step 2: Assume that the inequality holds for all natural n.

    Step 3: Check for n+1.
    We have
    u_{n+1}=-\frac{5}{u_n}+6

    If we sub. un for 1, we get
    u_{n+1}=-\frac{5}{1}+6 =1

    If we sub. un for 5, we get
    u_{n+1}=-\frac{5}{5}+6 =5

    Since we had assumed that 1<un<5, we conclude that 1<Un+1<5.
    This isn't valid reasoning: substituting two values for u_n (without any extra argument) doesn't prove anything about the behaviour for u_n between 1 and 5.

    Your argument should look more like:

    Since u_n < 5, we know that 5/u_n is ..., and so -5/u_n + 6 is ...
    Since u_n > 1, we know that 5/u_n is ..., and so -5/u_n + 6 is ...

    We conclude that 1 &lt; u_n+1 &lt; 5 as desired.
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    Thank you for your answers.Please could you tell me if this is correct?

    Step 3: Check for k+1

    u_{n+1} = -\frac{5}{u_n}+6
    >  -\frac{5}{u_n}
    < \frac{5}{u_n}

    We assumed that 1&lt;u_n&lt;5
    which implies that
    1&lt;\frac{5}{u_n}&lt;5

    Since u_{n+1}&lt;\frac{5}{u_n}, we conclude that 1&lt;u_{n+1}&lt;5
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    (Original post by razzor)
    Thank you for your answers.Please could you tell me if this is correct?

    Step 3: Check for k+1

    u_{n+1} = -\frac{5}{u_n}+6
    >  -\frac{5}{u_n}
    < \frac{5}{u_n}

    We assumed that 1&lt;u_n&lt;5
    which implies that
    1&lt;\frac{5}{u_n}&lt;5

    Since u_{n+1}&lt;\frac{5}{u_n}, we conclude that 1&lt;u_{n+1}&lt;5
    You have shown that u_{n+1}&lt;5, but you still need to show that 1&lt;u_{n+1}. The post by DFranklin highlights the way.
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    Please could you show me how to get there? I don't really understand his explanations
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    You have already done most of it.

    \frac{5}{u_n}&lt;5 \Rightarrow -\frac{5}{u_n}&gt;-5 \Rightarrow -\frac{5}{u_n} + 6&gt;-5+6 \Rightarrow u_{n+1}&gt;1
 
 
 
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