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 indeterminateform $\frac{0}{0}$ 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)