Given general term - find limits and comparison

[Saitama]

Homework Statement

Let $\displaystyle a_n=\frac 1 2+\frac 1 3+...+\frac 1 n$. Then

A)$a_n$ is less than $\displaystyle \int_2^n\frac{dx}{x}$.

B)$a_n$ is greater than $\displaystyle \int_1^n\frac{dx}{x}$.

C)$\displaystyle \lim_{n\rightarrow \infty} \frac{a_n}{\ln n}=1$

D)$\displaystyle \lim_{n\rightarrow \infty} a_n$ is finite.

Homework Equations

The Attempt at a Solution

To my knowledge, there is no known closed form for the given $a_n$.

I am clueless about the right approach so I started with $n=2$. For n=2, $a_2=0.5$

Also,
$$\int_2^2 \frac{dx}{x}=0$$
and
$$\int_1^2 \frac{dx}{x}=\ln 2 \approx 0.693$$
Obviously, A and B are not the answers.

How do I check for other options?

Any help is appreciated. Thanks!

 

[dirk_mec1]

D is not correct the series is diverging. C is basically saying that a(n) = ln(n) if n becomes large is that correct?

 

[Saitama]

D is not correct the series is diverging. C is basically saying that a(n) = ln(n) if n becomes large is that correct?

I don't have the answers at the moment, I will have them by tomorrow.

Can you please explain how do you get the series to be diverging?

Thanks!

 

[LCKurtz]

Think about comparing the series with approximating sums for the integrals in A and B.

 

[Saitama]

Think about comparing the series with approximating sums for the integrals in A and B.

I am not sure if I understand your statement but do you ask me this:
$$a_n=\int_2^n \frac{dx}{x}$$
?

 

[LCKurtz]

Think about comparing the series with approximating sums for the integrals in A and B.

I am not sure if I understand your statement but do you ask me this:
$$a_n=\int_2^n \frac{dx}{x}$$
?

Yes. Think about approximating that with rectangles and see if you can relate it to your series.

 

[Saitama]

Yes. Think about approximating that with rectangles and see if you can relate it to your series.

Thanks LCKrutz! :)

I just found the wiki page on the given series titled "Harmonic Number" and it contains a nice sketch of approximation using rectangles. http://en.wikipedia.org/wiki/Harmonic_number

Getting back to the question, as $n\rightarrow \infty$,
$$a_n=\ln n-\ln 2$$
$a_n$ is obviously not finite. That leaves us with option C.

$$\lim_{n\rightarrow \infty} \frac{a_n}{\ln n}=\lim_{n\rightarrow \infty} \frac{\ln n-\ln 2}{\ln n}=\lim_{n\rightarrow \infty} 1-\frac{\ln 2}{\ln n}=1$$
Hence, C is correct, thanks a lot LCKurtz!

 

[LCKurtz]

What is the exact statement of the problem? Is it a True-False type question?

[Edit] I see you answered that question while I was posting it.

 

[Saitama]

What is the exact statement of the problem? Is it a True-False type question?

What I wrote is the exact wording of the problem statement but I should have been more clear. The problem is from a test paper belonging to the section which consists of multiple choice questions. I have to select the correct options.