Problems from today calculus test

  • Thread starter twoflower
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  • #1
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The exam is behind, but I'll have to repeat it at least once :-)

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

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

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:

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

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.
 

Answers and Replies

  • #2
arildno
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[tex]log(1+\epsilon)\approx\epsilon-\frac{\epsilon^{2}}{2},\epsilon<<1[/tex]
[tex]\sin\epsilon\approx\epsilon-\frac{\epsilon^{3}}{6},\epsilon<<1[/tex]
Use these approximations to show that your first expression tends to [tex]-\frac{1}{2}[/tex]
 
  • #3
Galileo
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twoflower said:
[tex]
\sum_{n = 1}^{+\infty} (-1)^{n} \arctan \left( \sqrt{n^2 + 1} - \sqrt{n^2 - 1} \right)
[/tex]

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)...

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).
 
  • #4
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arildno said:
[tex]log(1+\epsilon)\approx\epsilon-\frac{\epsilon^{2}}{2},\epsilon<<1[/tex]
[tex]\sin\epsilon\approx\epsilon-\frac{\epsilon^{3}}{6},\epsilon<<1[/tex]
Use these approximations to show that your first expression tends to [tex]-\frac{1}{2}[/tex]

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:
  • #5
learningphysics
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twoflower said:
The exam is behind, but I'll have to repeat it at least once :-)

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

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

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

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:

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

Use L'Hopital's rule twice, and you get the answer = -1/2
 
  • #6
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Hi...I found your first problem rather interesting because I encountered a "somewhat" similar problem (http://www.jee.iitb.ac.in/maths/images/MQNo_05.gif [Broken]).

Cheers
Vivek
 
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