Problems from today calculus test

[twoflower]

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.

 

[arildno]

$$log(1+\epsilon)\approx\epsilon-\frac{\epsilon^{2}}{2},\epsilon<<1$$
$$\sin\epsilon\approx\epsilon-\frac{\epsilon^{3}}{6},\epsilon<<1$$
Use these approximations to show that your first expression tends to $$-\frac{1}{2}$$

 

[Galileo]

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

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)...

If $\arctan \left( \sqrt{n^2 + 1} - \sqrt{n^2 - 1} \right)$ 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).

 

[twoflower]

$$log(1+\epsilon)\approx\epsilon-\frac{\epsilon^{2}}{2},\epsilon<<1$$
$$\sin\epsilon\approx\epsilon-\frac{\epsilon^{3}}{6},\epsilon<<1$$
Use these approximations to show that your first expression tends to $$-\frac{1}{2}$$

Hmmm, I'm gallows-ripe. Now I found out why I wasn't able to solve this problem. In exercise book I have written the expansion of sin in a wrong way (without 'x' at the beginning)...

 

[learningphysics]

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...

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

 

[maverick280857]

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 ).

Cheers
Vivek