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Math Help - Monotonicity of a sequence

  1. #1
    Junior Member gusztav's Avatar
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    Monotonicity of a sequence

    Hi everyone,
    I need to determine whether the sequence (a_n), where a_n=\frac{<br />
\sqrt{n^2+1}-n}<br />
{\sqrt{n+3}+2n}, is decreasing or increasing. To do that, I can first suppose that it's e.g. increasing, and then see if the relation a_n \leq a_{n+1} holds for every n \in \mathbb{N} . In other words, if the inequality \frac{<br />
\sqrt{n^2+1}-n}<br />
{\sqrt{n+3}+2n}\leq<br />
\frac{\sqrt{(n+1)^2+1}-(n+1)}<br />
{\sqrt{(n+1)+3}+2(n+1)} is true, then is also a_n \leq a_{n+1}, and if it isn't, then a_n > a_{n+1} and the sequence is decreasing. This is where I got stuck, for no matter how I tried to solve this inequality (squaring both sides; multiplying whole inequality by both denominators etc), I can't seem to arrive at the solution.


    To give you a better idea of the way I am supposed to solve this, here is an easier example:
    determine whether the sequence (b_n), where b_n=\frac{2n+5}{5-7n}, is decreasing or increasing.
    We can start by assuming that (b_n) is increasing; in that case b_n \leq b_{n+1} is true if and only if \frac{2n+5}{5-7n} \leq \frac{2(n+1)+5}{5-7(n+1)} \Leftrightarrow ... \Leftrightarrow -14n^2-39n-10 \leq -14n^2-39n+35 \Leftrightarrow -10 \leq 35, which is true, and therefore the starting assumption ( b_n \leq b_{n+1}) is also true, and (b_n) is an increasing sequence.


    So, basically, I should somehow arrive from this expression: \frac{<br />
\sqrt{n^2+1}-n}<br />
{\sqrt{n+3}+2n}\leq<br />
\frac{\sqrt{(n+1)^2+1}-(n+1)}<br />
{\sqrt{(n+1)+3}+2(n+1)} to a statement from which the verity of the starting assumption (i.e. a_n \leq a_{n+1}) could be inferred.

    Many thanks.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by gusztav View Post
    Hi everyone,
    I need to determine whether the sequence (a_n), where a_n=\frac{<br />
\sqrt{n^2+1}-n}<br />
{\sqrt{n+3}+2n}, is decreasing or increasing. To do that, I can first suppose that it's e.g. increasing, and then see if the relation a_n \leq a_{n+1} holds for every n \in \mathbb{N} . In other words, if the inequality \frac{<br />
\sqrt{n^2+1}-n}<br />
{\sqrt{n+3}+2n}\leq<br />
\frac{\sqrt{(n+1)^2+1}-(n+1)}<br />
{\sqrt{(n+1)+3}+2(n+1)} is true, then is also a_n \leq a_{n+1}, and if it isn't, then a_n > a_{n+1} and the sequence is decreasing. This is where I got stuck, for no matter how I tried to solve this inequality (squaring both sides; multiplying whole inequality by both denominators etc), I can't seem to arrive at the solution.


    To give you a better idea of the way I am supposed to solve this, here is an easier example:
    determine whether the sequence (b_n), where b_n=\frac{2n+5}{5-7n}, is decreasing or increasing.
    We can start by assuming that (b_n) is increasing; in that case b_n \leq b_{n+1} is true if and only if \frac{2n+5}{5-7n} \leq \frac{2(n+1)+5}{5-7(n+1)} \Leftrightarrow ... \Leftrightarrow -14n^2-39n-10 \leq -14n^2-39n+35 \Leftrightarrow -10 \leq 35, which is true, and therefore the starting assumption ( b_n \leq b_{n+1}) is also true, and (b_n) is an increasing sequence.


    So, basically, I should somehow arrive from this expression: \frac{<br />
\sqrt{n^2+1}-n}<br />
{\sqrt{n+3}+2n}\leq<br />
\frac{\sqrt{(n+1)^2+1}-(n+1)}<br />
{\sqrt{(n+1)+3}+2(n+1)} to a statement from which the verity of the starting assumption (i.e. a_n \leq a_{n+1}) could be inferred.

    Many thanks.
    hmm... i'd use induction (though it would probably be messy here). find a_1 and a_2 and base your assumption on those values (that is, if a_1 \ge a_2 then assume it is decreasing, otherwise assume it is increasing). then try to prove the rest by induction. do you know how to do mathematical induction?
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  3. #3
    Junior Member gusztav's Avatar
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    Quote Originally Posted by Jhevon View Post
    hmm... i'd use induction (though it would probably be messy here). find a_1 and a_2 and base your assumption on those values (that is, if a_1 \ge a_2 then assume it is decreasing, otherwise assume it is increasing). then try to prove the rest by induction. do you know how to do mathematical induction?
    Yes. It's a great idea, thanks, I'll try it.
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