Re: Limit+fractional part

In article
<1151679541.612002.284610@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"eugene" <jane1806@xxxxxxxxxx> wrote:

Patrick Coilland wrote:
eugene nous a récemment amicalement signifié :
A N Niel wrote:
In article <1151671442.146228.172730@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
eugene <jane1806@xxxxxxxxxx> wrote:

Here is the limit:

limt {n -> oo} sqrt(N) * ( 1 - max {1 <= n <= N} ( sqrt{n} ) ) .

I've got 0 as the value of this limit. Am i right?

Thanks


Your n and N seem confused...

Yes, you are right. Sorry

Here is the limit:

limt {N -> oo} sqrt(N) * ( 1 - max {1 <= n <= N} ( sqrt{n} ) ) .

I've got 0 as the value of this limit. Am i right?


I don't understand :
What is "sqrt{n}" ?
Is it the same as "sqrt(n)" ?

This sqrt{n} is the same as sqrt(n), but this

sqrt{n} is in the brackets { ... } which mean the fractional part.

***. { ... } mean the fractional part.

Assuming you mean frac(sqrt(n)), let

a_N = sqrt(N) *(1 - max{frac(sqrt(n)) : 1 <= n <= N})

If N = m^2, the max above occurs at n = m^2 - 1, where it's
easily calculated: 1 - 1/[m + sqrt(m^2-1)]. So if m^2 <= k <
(m+1)^2, then

m/[(m+1)+sqrt((m+1)^2-1)] <= a_k <= (m+1)/[m+sqrt(m^2-1)].

The upper and lower estimates both -> 1/2 as m -> oo, so the
limit is 1/2, as Patrick stated.

PS: You need to state your problems more carefully! It's not that
hard to do.
.