# example using Stolz-Cesaro theorem

Example: We try to determine the value of

 $\lim_{n\to\infty}\frac{1^{k}+2^{k}+...+n^{k}}{n^{k+1}},\,k\in\mathbb{N}.$

We consider the sequences $\alpha_{n\geq 1}=1^{k}+2^{k}+...+n^{k}$ and $\beta_{n\geq 1}=n^{k}$ and using the Stolz-Cesaro theorem we have that

 $\displaystyle\lim_{n\to\infty}\frac{1^{k}+2^{k}+...+n^{k}}{n^{k+1}}=$ (1) $\displaystyle\lim_{n\to\infty}\frac{(1^{k}+2^{k}+...+(n+1)^{k})-(1^{k}+2^{k}+.% ..+n^{k})}{(n+1)^{k+1}-n^{k+1}}=$ (2) $\displaystyle\lim_{n\to\infty}\frac{(n+1)^{k}}{{(n+1)^{k+1}-n^{k+1}}}.$ (3)

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

 $\displaystyle\lim_{n\to\infty}\frac{1}{(n+1)-n^{k+1}(n+1)^{-k}}=$ (4) $\displaystyle\lim_{n\to\infty}\frac{1}{n(1+n^{-1}-(1+n^{-1})^{-k})}=$ (5) $\displaystyle\lim_{n\to\infty}\frac{n^{-1}}{1+n^{-1}-(1+n^{-1})^{-k}}.$ (6)

By applying L’Hôpital’s rule once we get

 $\displaystyle\lim_{n\to\infty}\frac{n^{-1}}{1+n^{-1}-(1+n^{-1})^{-k}}=$ (7) $\displaystyle\lim_{n\to\infty}\frac{-n^{-2}}{-n^{-2}-k(1+n^{-1})^{-k-1}n^{-2}}=$ (8) $\displaystyle\lim_{n\to\infty}\frac{1}{1+k(1+n^{-1})^{-k-1}}=$ (9) $\displaystyle\frac{1}{1+k}.$ (10)
Title example using Stolz-Cesaro theorem ExampleUsingStolzCesaroTheorem 2013-03-22 15:31:02 2013-03-22 15:31:02 georgiosl (7242) georgiosl (7242) 4 georgiosl (7242) Example msc 40A05 StolzCesaroTheorem LHpitalsRule