# Math Help - Find the Limit: Urgent. (1^p+2^p+.....+n^p) /(n^(p +1))

1. ## Find the Limit: Urgent. (1^p+2^p+.....+n^p) /(n^(p +1))

Hi All,

I'm unable to find the exact steps to find out the answer of the following q?

(1^p+2^p+.....+n^p)
Limit ---------------------
n->~(infinity) (n^(p+1))

Arun Sangal
91-9811514096
arun.sangal@galileo.com

2. Originally Posted by sangal_ak04
Hi All,

I'm unable to find the exact steps to find out the answer of the following q?

(1^p+2^p+.....+n^p)
Limit ---------------------
n->~(infinity) (n^(p+1))

Arun Sangal
91-9811514096
arun.sangal@galileo.com
There is a general formula to add,
$1^p+2^p+...+n^p = A_{p+1}n^{p+1}+A_pn^p+...+A_1+A_0$
The coefficients follow Bernoulli numbers and get messy.
However the first ceofficient is simply $A_{p+1}=\frac{1}{(p+1)}$.

So we have,
$\frac{\frac{1}{(p+1)}n^{p+1}+A_pn^p+...+A_1p+A_0}{ n^{p+1}}$.
And this limit is therefore,
$\boxed{ \frac{1}{(p+1)} }$.

3. Use the Stolz-Cesaro criteria.
Let $a_n=1^p+2^p+\ldots+n^p, \ b_n=n^{p+1}$.
$(b_n)_{n\in\mathbf{N}}$ is ascendind and unbounded.
We calculate $\displaystyle\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lim_{n\to\infty}\frac{(n+1)^p}{(n+1)^{p+1}-n^{p+1}}=$
$\displaystyle =\lim_{n\to\infty}\frac{n^p+C_n^1n^{p-1}+\ldots}{n^{p+1}+C_{p+1}^1n^p+\ldots -n^{p+1}}=\frac{1}{p+1}$

So $\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}=\fra c{1}{p+1}$.