Let (xn)n≥1 defined as follows:
x1>0,x1+x2+⋯+xn=1√xn+1.
Compute the limit limn→∞n2x3n.
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 n≥1. Then, the recursion reads y2n(yn+1−yn)=1. We see that yn+1−yn>0 for n≥1. Since y2 is monotone, it follows that
n=n∑k=1y2k(yk+1−yk)≤∫yn+1y1y2dy=13(y3n+1−y31),
i.e. yn+1≥31/3n1/3.
Consequently 1y2n≤3−2/31(n−1)2/3,
and yn+1=y2+n∑k=21y2k≤y2+3−2/3n∑k=21(k−1)2/3=y2+3−2/3n−1∑k=11k2/3.
Now we can estimate the RHS with a telescope: k1/3−(k−1)1/3=1k2/3+k1/3(k−1)1/3+(k−1)2/3≥13k2/3,
that means yn+1≤y2+31/3(n−1)1/3.
(1) and (2) together give limn→∞ynn−1/3=31/3.
Since xn+1=yn+1−yn=1/y2n, we have limn→∞xnn2/3=3−2/3,
and thus limn→∞x3nn2=3−2=19.
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