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Direct Comparison Test

  1. Apr 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Use any test to determine whether the series converges.


    2. Relevant equations
    [itex]\displaystyle \sum^{∞}_{n=1} tan(1/n) [/itex]


    3. The attempt at a solution
    Direct Comparison Test

    tan(1/n) > 1/n

    By integral test: 1/n diverges thus, by dct, tan(1/n) diverges.
     
  2. jcsd
  3. Apr 4, 2013 #2

    BruceW

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    looks good to me. They might want some more detail in your working, i.e. what is the range of values that 1/n can take. And why is tan(1/n) > 1/n for these range of values?
     
  4. Apr 4, 2013 #3

    Zondrina

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    Yes, comparison test is what you want here. You may want to note that you need sufficiently large n for it to hold.
     
  5. Apr 4, 2013 #4

    LCKurtz

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    As others have noted, you need to say why this is true.

    It isn't 1/n that diverges, it is ##\sum 1/n## that diverges. Same for tan(1/n).
     
  6. Apr 4, 2013 #5
    i thought it held only for sufficiently small n. it holds starting at n=1
     
  7. Apr 4, 2013 #6

    BruceW

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    I think you are right that it holds, starting at n=1. But think about what would happen if you started at smaller n, for example starting at n=0.0001, would it still hold then?

    Edit: and think about if it started at n=1000, would it hold? This should suggest whether it holds for sufficiently small or sufficiently large n.

    Edit again: hint: look at the graph of tan(x), where might there be problems for the equation tan(x)>x ? And then think about how this applies to n (which is 1/x)
     
    Last edited: Apr 4, 2013
  8. Apr 4, 2013 #7

    LCKurtz

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    Think about the graphs of y = tan(x) and y = x for nonnegative x.
     
  9. Apr 4, 2013 #8

    Mark44

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    Why would you want to check n = 0.0001? n is an index on a summation, typically limited to integer values.
     
  10. Apr 4, 2013 #9
    exactly.
     
  11. Apr 4, 2013 #10
    It it doesn't seem that hard to show that [itex] \tan \frac{1}{n} > \frac{1}{n} [/itex] for all positive n. Since [itex] \frac{1}{n} \leq 1 [/itex], it suffices to show that [itex] \tan x \geq x [/itex] for [itex] 0 \leq x \leq 1 [/itex] . Just let [itex] f(x) = \tan x - x [/itex] and find [itex] f'(x) = \sec^2 x -1 \geq 0[/itex]. Since [itex] f [/itex] is zero at x=0 and increasing, it must be positive on [itex] [0,1] [/itex] as it is continous. If [itex] \tan x - x [/itex] is positive, then [itex] \tan x \geq x [/itex] and we are done.
     
  12. Apr 4, 2013 #11

    BruceW

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    @HS-Scientist: yeah, don't give the game away though... (p.s. is this a British expression, or does this make sense to Americans, too?)

    It doesn't have to be integer though. The point is that whatlifeforme was saying that n needs to be sufficiently small, but actually it holds for n sufficiently large. Although I admit, checking particular values of n is not very helpful. It is better to look at the graph.
     
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