example using Stolz-Cesaro theorem

Example: We try to determine the value of

$$\underset{n\to \mathrm{\infty }}{lim}\frac{1^k+2^k+\mathrm{\dots }+n^k}{n^{k+1}},k\in \mathbb{N}.$$

We consider the sequences ${\alpha }_{n\ge 1}=1^k+2^k+\mathrm{\dots }+n^k$ and ${\beta }_{n\ge 1}=n^k$ and using the Stolz-Cesaro theorem we have that

$\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{1^k+2^k+\mathrm{\dots }+n^k}{n^{k+1}}}=$ (1)

$\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{(1^k+2^k+\mathrm{\dots }+{(n+1)}^k)-(1^k+2^k+\mathrm{\dots }+n^k)}{{(n+1)}^{k+1}-n^{k+1}}}=$ (2)

$\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{{(n+1)}^k}{{(n+1)}^{k+1}-n^{k+1}}}.$ (3)

Now we try to get the expression in the indeterminate form $\frac{0}{0}$ as n approaches $\mathrm{\infty }$, dividing numerator and denominator of (3) by ${(n+1)}^k$.

$\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{(n+1)-n^{k+1}{(n+1)}^{-k}}}=$ (4)

$\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{n(1+n^{-1}-{(1+n^{-1})}^{-k})}}=$ (5)

$\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{n^{-1}}{1+n^{-1}-{(1+n^{-1})}^{-k}}}.$ (6)

By applying L’Hôpital’s rule once we get

$\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{n^{-1}}{1+n^{-1}-{(1+n^{-1})}^{-k}}}=$ (7)

$\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{-n^{-2}}{-n^{-2}-k{(1+n^{-1})}^{-k-1}{n}^{-2}}}=$ (8)

$\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{1+k{(1+n^{-1})}^{-k-1}}}=$ (9)

${\displaystyle \frac{1}{1+k}}.$ (10)

Title	example using Stolz-Cesaro theorem
Canonical name	ExampleUsingStolzCesaroTheorem
Date of creation	2013-03-22 15:31:02
Last modified on	2013-03-22 15:31:02
Owner	georgiosl (7242)
Last modified by	georgiosl (7242)
Numerical id	4
Author	georgiosl (7242)
Entry type	Example
Classification	msc 40A05
Related topic	StolzCesaroTheorem
Related topic	LHpitalsRule