Sunday, 12 January 2014

real analysis - Convergent sequence and limlimitsntoinftyn(an+1an)=0




Let (an)n be a real, convergent, monotonic sequence. Prove that if the limit lim exists, then it equals 0.





I tried to apply the Stolz-Cesaro theorem reciprocal:
\lim_{n\to \infty}n(a_{n+1}-a_n)=\lim_{n \to \infty} \frac{a_n}{1+\frac{1}{2}+\dots+\frac{1}{n-1}}=0 but I can't apply it since for b_n=1+\frac{1}{2}+\dots+\frac{1}{n-1} we have \lim_\limits{n \to \infty} \frac{b_{n+1}}{b_n}=1. I also attempted the \epsilon proof but my calculations didn't lead to anything useful.


Answer



Hint. You are on the right track. Note that a_n is convergent to a finite limit and the harmonic series at the denominator is divergent. Therefore
0=\lim_{n \to \infty} \frac{a_n}{1+\frac{1}{2}+\dots+\frac{1}{n-1}}\stackrel{\text{SC}}{=}\lim_{n\to \infty}\frac{a_{n+1}-a_n}{\frac{1}{n}}=\lim_{n\to \infty}n(a_{n+1}-a_n).


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