1. Nov 12, 2008

### Diffy

1. The problem statement, all variables and given/known data

Two questions

lim as n -> infinity

(1^k + 2^k + ... + n^k) / n^(k+1)

and

lim as n -> infinity

1/(n+1) + 1/(n+2) + ... + 1/2n

2. Relevant equations
Definitions of limits, laws of exponents etc.

3. The attempt at a solution

Well I think I have them both solved but it seems too easy therefore I think I did something wrong or am missing something.

I think they are both zero.

The first one you can rewrite the term in the limit as

(1^k)/(n^(k+1)) + ... + (n^k)/(n^(k+1))

If you take the limit as n goes to infinity of each of these, then they all go to zero.

I am not quite sure I can do this, or argue this that way, but it is the only thing that comes to mind.

For the second question, I do the same thing. Take the limit as n goes to infinity of each term. Since they all go to 0 the entire limit goes to zero.

Please let me know if my reasoning is correct.

2. Nov 12, 2008

### Staff: Mentor

One quick santiy check -- use Excel to do a few examples of each summation...

Start with n=10, then cut and paste to make n=20, etc. Do you see a trend?

3. Nov 12, 2008

### Staff: Mentor

BTW, you will need a column for n, and a column for the numerator of the nth term, and a column for the denominator of the nth term, and a column for the quotient of the nth term, and then a sum over that last column down at the bottom.

4. Nov 12, 2008

### Diffy

Yea they don't look to be zero, the first one looks like 1/3 and the second one looks like... well I don't know with n = 200 it was like .69 and n = 300 it was over 1.

Can you please give me a hint on how to actually solve these guys?

5. Nov 12, 2008

### Dick

You must have taken k=2 for the first one, right? They are both Riemann sums which become exact in the limit as n -> infinity. The first one is the integral from 0 to 1 of x^k. The second one is the integral from 1 to 2 of 1/x. The first one depends on k, but if k=2 it's 1/3. That's how I knew you used k=2. The second is ln(2)=0.693... Good job using Excel (yuck!). Very brave. Review Riemann sums and see how this works, ok?

6. Nov 12, 2008

### Diffy

Dick thank you very much for your response. I owe ya one, how about next time you need Excel help I'm your guy!

7. Nov 12, 2008

### Dick

Actually the once or twice in my life I've actually used a spreadsheet, I used Openoffice Calc. But I'm sure you can you can handle that as well. I'll keep you in mind.