Monday, 20 August 2018

sequences and series - Compute: limlimitsntoinftyfracxnlnn



Let be the sequence (xn)n0, x0 a real number, and defined as follows: xn+1=xn+exn


Compute the limit:



limnxnlnn




Luckily, I've found another post here with a very similar question. If you know other ways to solve it and want to share them then I'd be very grateful for that.


Answer



Heuristically, this difference equation resembles the differential equation x(t)=ex, which has solutions x(t)=ln(t+C), so I would expect the answer to be 1.



In fact, let xn=ln(n+sn), and y(t)=ln(t+sn). Then for nt<n+1 we have
y(t)>y(n)=xn, so
ln(n+1+sn)ln(n+sn)=n+1ndtt+sn=n+1ney(t) dt<n+1nexn dt=exn=xn+1xn



i.e. xn+1ln(n+1+sn)>xnln(n+sn)=0, or sn+1>sn.
Thus xn>ln(n+s0). But then xn+1xn=exn<1n+s0
so xn<x0+n1j=01j+s0. Since both upper and lower bounds are ln(n)+O(1), we conclude that xn=ln(n)+O(1).


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