# Limit with parameter

1. Dec 10, 2004

### Nobody1111

Find limit n-->infinity of sequence a_n:

a_n = (1^k+2^k+...+n^k)/(n^(k+1)), where k is parameter.

IThanks from advance for any help.

I tried to compute this limit using Stolz Theorem, but I don't know if I can do it in this way.

2. Dec 11, 2004

### learningphysics

What value can k take? The analysis below is for k being an integer and k>=0. For k<0 the denominator is not strictly increasing, so Stolz theorem doesn't hold.

For k>=0 and k being an integer, I think Stolz theorem is a good way to solve the problem:

By applying it... try to find the limit as n->infinity of:

$$\frac{(n+1)^k}{(n+1)^{(k+1)}-n^{(k+1)}}$$

You can factor the denominator:

$$(n+1)^{(k+1)}-n^{(k+1)}=[(n+1)-n][(n+1)^k +(n+1)^{(k-1)}n+(n+1)^{(k-2)}n^2+...]$$

Then if you divide both the numerator and denominator by $$(n+1)^k$$ it should be easy to see the limit. Hint: the answer depends on k.

Last edited: Dec 11, 2004
3. Dec 12, 2004

### Nobody1111

One more question: What if k=-1. The limit doesn't exist or limit equals to infinity?

4. Dec 12, 2004

### learningphysics

Yes, the limit does not exist (or another way to say it is the limit is infinity).

If you plug in k=-1 into your sequence you get:

$$a_n=\begin{array}{c}n\\\sum\\m=1\end{array}1/m$$

Since $$\sum1/m$$ is a divergent series, as n->infinity a_n->infinity.