
Re: Analysis with lim a_0.sqrt(n)
Posted:
Apr 11, 2013 4:05 AM


On Thu, 11 Apr 2013, mina_world@hanmail.net wrote:
> Constant a_0, a_1, a_2, ... , a_k > > a_0 + a_1 + a_2 + ... + a_k = 0 > > Find the limit, > > lim{n>oo}{(a_0)(sqrt(n)) + (a_1)(sqrt(n+1)) + (a_2)(sqrt(n+2)) + ... > > + (a_k)(sqrt(n+k))} > > Hm... I need your advice. When stumped, do the easy ones.
If k = 0, the limit is zero. Trivial If k = 1, the limit is zero. a0 + a1 = 0 lim(n>oo) (a0.sqr n + a1.sqr n+1) = a0,lim(n>oo) (sqr n  sqr n+1) = a0.lim(n>oo) 1/(sqr n + sqr n+1) = 0 Is the limit is zero?
sum(j=0,k+1) aj.sqr n+j = sum(j=0,k) aj.sqr n+j + a_(j+1) sqr n+j+1 = sum(j=0,k1) aj.sqr n+j  (sum(k1) aj)sqr n+k + sum(k=0,k) aj.sqr n+j + a_(k+1) sqr n+j+1 = a_(k+1) sqr n+j + a_(k+1) sqr n+j+1
Yes, as before, the limit is zero.

