example using Stolz-Cesaro theorem


Example: We try to determine the value of

limn1k+2k++nknk+1,k.

We consider the sequences αn1=1k+2k++nk and βn1=nk and using the Stolz-Cesaro theorem we have that

limn1k+2k++nknk+1= (1)
limn(1k+2k++(n+1)k)-(1k+2k++nk)(n+1)k+1-nk+1= (2)
limn(n+1)k(n+1)k+1-nk+1. (3)

Now we try to get the expression in the indeterminate form 00 as n approaches , dividing numerator and denominator of (3) by (n+1)k.

limn1(n+1)-nk+1(n+1)-k= (4)
limn1n(1+n-1-(1+n-1)-k)= (5)
limnn-11+n-1-(1+n-1)-k. (6)

By applying L’Hôpital’s rule once we get

limnn-11+n-1-(1+n-1)-k= (7)
limn-n-2-n-2-k(1+n-1)-k-1n-2= (8)
limn11+k(1+n-1)-k-1= (9)
11+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