Monday, 24 November 2014

real analysis - Limit of bn when bn=fracann and limlimitsntoinfty(an+1an)=l




I have the following problem.



I have to find the limit of bn=ann, where lim



My approach:



I express a_n in terms of b_n, i.e.
a_n=nb_n and a_{n+1}=(n+1)b_n




We look at the difference: a_{n+1}-a_n=(n+1)b_{n+1}-nb_n



Assuming that b_n converges to a real number m, we see that:



l=(n+1)m-nm, from where I conclude that m=l.



What I'm left with is proving that b_n is convergent which I'm not sure how to do.



Thanks in advance!



Answer



As @ParamanandSingh suggested, from Stolz–Cesàro theorem
a_{n+1}-a_n=\frac{a_{n+1}-a_n}{(n+1)-n} \rightarrow l, n \rightarrow \infty
where \{n\}_{n \in \mathbb{N}} is monotone and divergent, then
\frac{a_n}{n} \rightarrow l, n \rightarrow \infty


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