Monday, 22 September 2014

limits - Sequence defined recursively: sumnk=1xk=frac1sqrtxn+1




Let (xn)n1 defined as follows:
x1>0,x1+x2++xn=1xn+1.

Compute the limit limnn2x3n.



MY TRY: I thought about using Stolz-Cesaro lemma, but I couldn't get to an appropriate form that leads to an easier limit.


Answer



Let yn=x1++xn for n1. Then, the recursion reads y2n(yn+1yn)=1. We see that yn+1yn>0 for n1. Since y2 is monotone, it follows that
n=nk=1y2k(yk+1yk)yn+1y1y2dy=13(y3n+1y31),


i.e. yn+131/3n1/3.
Consequently 1y2n32/31(n1)2/3,
and yn+1=y2+nk=21y2ky2+32/3nk=21(k1)2/3=y2+32/3n1k=11k2/3.
Now we can estimate the RHS with a telescope: k1/3(k1)1/3=1k2/3+k1/3(k1)1/3+(k1)2/313k2/3,
that means yn+1y2+31/3(n1)1/3.

(1) and (2) together give limnynn1/3=31/3.


Since xn+1=yn+1yn=1/y2n, we have limnxnn2/3=32/3,
and thus limnx3nn2=32=19.


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...