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anonymous

  • 5 years ago

Prove that (n^2)/ (1+n^2) is decreasing, in other words, that... ((n+1)^2/(1+(n+1)^2)) is less than (n^2/(1+n^2))

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  1. anonymous
    • 5 years ago
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    Are you sure this is decreasing? Because the only way to show a sequence is decreasing is to do the following: \[a_{n+1} \le a_{n}\] This can be shown in 4 other ways \[a_{n+1} - a_{n} \le 0 \]\[a_{n} - a_{n+1} \ge 0\]\[a_{n+1}/a_{n} \le 1\]\[a_{n}/a_{n+1} \ge 1\] I did the last one and I couldn't show that the sequence was decreasing. Here's what I got: \[(n+1)^{2}/1+(n+1)^{2} \le n^{2}/(1+n^{2})\]

  2. anonymous
    • 5 years ago
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    Ugh... sorry, accidentally pressed post. start from the last inequality \[(n+1)^{2}(1+n^{2}) \le n^{2}(1+(n+1)^{2})\]\[(n+1)^{2}+n^{2}(n+1)^{2} \le n^{2} + n^{2}(n+1)^{2}\]\[n^{2}[(n+1)^{2}/n^{2} + (n+1)^{2}] \le n^{2}(1+(n+1)^{2})\]\[(n+1)^{2}/n^{2}+(n+1)^{2} \le 1 + (n+1)^{2}\]\[(n+1)^{2}/n^{2} \le 1\]\[(n+1)^{2} \le n^{2}\]And it's fairly obvious this isn't true. So it's got to be an increasing sequence unless I did something wrong.

  3. anonymous
    • 5 years ago
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    so since it is increasing, the work u did above proves it? your work does make sense. ihonestlydont know why my teacher said that it was decreasing.

  4. anonymous
    • 5 years ago
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    I'm pretty sure that's correct.

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