Finding Limit: Homework Statement & Solution

[diredragon]

Homework Statement

Find the limit of the following sequence:
$L_2 = \lim_{n \rightarrow +\infty} \frac {\sum_{k=0}^n (2k - 1)^p}{n^{p+1}}$

Homework Equations

3. The Attempt at a Solution [/B]
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 I am stuck on how to further simplify this. Could get a hint from someone that knows?

 

[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.

 

[diredragon]

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.

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

 

[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 !

 

[BvU]

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

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

 

[diredragon]

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 !

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?

 

[BvU]

It is. Well done !

 

[andrewkirk]

@diredragon You're welcome.

 

[diredragon]

@diredragon You're welcome.

Yeah thanks, you provided a key hint :)

 

[Ray Vickson]

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?

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
$$\lim_{n \to \infty} \frac{a(n)}{b(n)} = 1.$$

 

[diredragon]

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
$$\lim_{n \to \infty} \frac{a(n)}{b(n)} = 1.$$

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