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Problems from today calculus test

  1. Jan 19, 2005 #1
    The exam is behind, but I'll have to repeat it at least once :-)

    Here are two problems I wasn't able to solve:

    \lim_{n \rightarrow \infty} n^2 \left[ \log \left( 1 + \frac{1}{n} \right) - \sin \left( \frac{1}{n} \right) \right]

    I tried to solve it using Taylor, but it didn't help me...

    And the second one, which I didn't even try, because I didn't catch it:

    Convergence and absolute convergence of this:

    \sum_{n = 1}^{+\infty} (-1)^{n} \arctan \left( \sqrt{n^2 + 1} - \sqrt{n^2 - 1} \right)

    How should I do that? IMO it would be sufficient that the arctan goes to 0 and then the sum would converge (Leibniz's rule)...

    Thank you.
  2. jcsd
  3. Jan 19, 2005 #2


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    Dearly Missed

    Use these approximations to show that your first expression tends to [tex]-\frac{1}{2}[/tex]
  4. Jan 19, 2005 #3


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    If [itex]\arctan \left( \sqrt{n^2 + 1} - \sqrt{n^2 - 1} \right)[/itex] decreases monotonically to zero, then the series converges. (It does)
    This is by the convergence criterium of Dirichlet, or a special case of it caled the Alternating Series test. It's probably the same rule you call Leibniz' rule. (He's already got so many rules with his name).
  5. Jan 19, 2005 #4
    Hmmm, I'm gallows-ripe. Now I found out why I wasn't able to solve this problem. In excercise book I have written the expansion of sin in a wrong way (without 'x' at the beginning)...
    Last edited: Jan 19, 2005
  6. Jan 19, 2005 #5


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    For these type of problems I always find variable substitution helpful. I'll assume the log is log base e. Is that right?

    Let a=1/n , so the limit becomes:

    \lim_{a \rightarrow 0} \frac {\left[ \log \left( 1 + a \right) - \sin \left( a \right) \right]} {a^2}

    Use L'Hopital's rule twice, and you get the answer = -1/2
  7. Jan 20, 2005 #6
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