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Topic: Analysis with lim a_0.sqrt(n)
Replies: 5   Last Post: Apr 13, 2013 11:50 AM

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William Elliot

Posts: 1,437
Registered: 1/8/12
Re: Analysis with lim a_0.sqrt(n)
Posted: Apr 11, 2013 4:05 AM
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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,k-1) aj.sqr n+j - (sum(k-1) 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.



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