# Limit of this sequence.

1. Aug 22, 2011

### Oster

I have a sequence {xn} defined by

xn = 1/n[1 + 1/2 + 1/3 + ... + 1/n]
for all natural numbers n.

I want to show that this sequence converges to 0, i.e. given any positive real number 'r', I want to show that there exists a natural number k such that xk < r. (The sequence is monotonically decreasing.)

2. Aug 22, 2011

### Dick

Find an upper bound for [1 + 1/2 + 1/3 + ... + 1/n] using an integral to estimate it. That would be a good start.

Last edited: Aug 22, 2011
3. Aug 22, 2011

### l'Hôpital

Alternatively, you could prove the general case:
$$a_n \rightarrow a \Rightarrow \frac{1}{n} \displaystyle \sum_{k=1}^{n} a_k \rightarrow a$$

4. Aug 22, 2011

### hunt_mat

How about an application of the sandwich theorem.

5. Aug 23, 2011

### Oster

Hi!
[1 + 1/2 + ... + 1/n] is always less than [1+ln(n)] ?
implying that my sequence is less that [1/n + (1/n)*ln(n)] which I believe converges to 0.

6. Aug 23, 2011

### Dick

All true. But how did you show the sum is less than 1+ln(n)? And 'believing' [1/n + (1/n)*ln(n)] converges to zero doesn't prove it does. How would you show that?

7. Aug 24, 2011

### Oster

Isn't it something like the integral test? It was kinda obvious from the diagram.
And lim [x^x] -> 1 as x->0 follows from L'Hopital's rule. (The log of the limit is 0)

8. Aug 24, 2011

### GreenPrint

you could use the divergence test

9. Aug 24, 2011

### Dick

Yes, it is l'Hopital, but I don't see what x^x has to do with it. And yes, the integral part is sort of obvious from a lower sum estimate of 1/x. If you don't want to actually spell out the details, that's fine with me.