Finding the limit

Tags:
1. Jun 14, 2017

diredragon

1. The problem statement, all variables and given/known data
Find the limit of the following sequence:
$L_2 = \lim_{n \rightarrow +\infty} \frac {\sum_{k=0}^n (2k - 1)^p}{n^{p+1}}$
2. Relevant equations
3. The attempt at a solution

Seeing that $\lim_{n \rightarrow +\infty} n^{p+1} = + \infty$ i can apply the Stolz theorem. (Is something more necessary alongside it being divergent?).
The upper sum when used in Stolz theorem yields just $(2n+1)^{p}$ because the rest of the sum is subtracted in form $(n+1) - (n)$. I get

$\lim_{n \rightarrow +\infty} \frac{(2n+1)^p}{(n+1)^{p+1} - n^{p+1}}$
$\lim_{n \rightarrow +\infty} \frac{(2n+1)^p}{(n+1)^p + (n+1)^{p-1}n + (n+1)^{p-2}n^2 + ... + n^p}$
And im stuck on how to further simplify this. Could get a hint from someone that knows?

2. Jun 14, 2017

andrewkirk

i suggest as a next step dividing the numerator and denominator by $n^p$.
You should then be able to calculate separate limits for the numerator and denominator, which are neither infinity nor zero. Hence you can use the division law to calculate the limit of the ratio.

3. Jun 14, 2017

diredragon

I get $\frac{2^p}{p+1}$

4. Jun 14, 2017

BvU

Check your last step, e.g. with a simple example like p = 2. I am missing the binomial coefficients and the powers of n should be descending, not remain equal !

5. Jun 14, 2017

BvU

Me too. (from your one-but-last step )

6. Jun 14, 2017

diredragon

You are right. I did mess something up. Weird that i got the right result when i divided by $n^{p}$. So:
$(n+1)^{p+1} - n^{p+1} = n^{p+1} + {{p+1}\choose{1}}n^p + {{p+1}\choose{2}}n^{p-1} + ... + {{p+1}\choose{k}}n^{p-k} + ... +1 - n^{p+1}$
$(n+1)^{p+1} - n^{p+1} ={{p+1}\choose{1}}n^p + {{p+1}\choose{2}}n^{p-1} + ... + {{p+1}\choose{k}}n^{p-k} + ... + 1$
If i know divide with $n^p$ i will obtain the correct result. Is this correct?

7. Jun 14, 2017

BvU

It is. Well done !

8. Jun 14, 2017

andrewkirk

9. Jun 14, 2017

diredragon

Yeah thanks, you provided a key hint :)

10. Jun 15, 2017

Ray Vickson

Better:
$$(n+1)^{p+1}-n^{p+1} = n^{p+1}\left [ \left(1 +\frac{1}{n} \right)^{p+1} - 1 \right],$$
and use $(1+x)^{p+1} = 1 + px +O(x^2)$ for small $x = 1/n$. Thus
$$(n+1)^{p+1}-n^{p+1} \sim (p+1)n^p$$
for large $n$. Here, $\sim$ denotes "asymptotic equality": $a(n) \sim b(n)$ for large $n$ if
$$\lim_{n \to \infty} \frac{a(n)}{b(n)} = 1.$$

11. Jun 15, 2017

diredragon

That is a really interesting approach too. I think i like it the most :D