1. ## sequence

Let an be the n-th term of the following sequence 1/1, 1/4, 3/4, 1/9, 3/9, 5/9, 1/16, 3/16, 5/16, 7/16, 1/25...

Find the maximal n satisfying an >= 1/10.

I do not know how to ask that question

2. ## Re: sequence

Originally Posted by provasanteriores
Let an be the n-th term of the following sequence:
1/1, 1/4, 3/4, 1/9, 3/9, 5/9, 1/16, 3/16, 5/16, 7/16, 1/25...

Find the maximal n satisfying an >= 1/10.

I do not know how to ask that question
And I do not know WHY you ask that question!
"an" should be shown as a(n): ok?
And the denominators are squares: ok?

Terms : number of terms (number of terms so far)
1/1^2 : 1 (1)
1/2^2, 3/2^2 : 2 (3)
1/3^2, 3/3^2, 5/3^2 : 3 (6)
and so on...
So what you have is "sum of natural numbers" so far...

Denominator 13^2 ends with 19/13^2 which is the 220th term.
(1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 = 220).
So at 220th term, we have 19/13^2 = .11242....; still > 1/10.

The next bunch ends with 21/15^2 which is 286th term:
21/15^2 = .09333....

I'm not doing any more for you...got a headache!!

3. ## Re: sequence

Originally Posted by provasanteriores
Let an be the n-th term of the following sequence
1/1, 1/4, 3/4, 1/9, 3/9, 5/9, 1/16, 3/16, 5/16, 7/16, 1/25...
Find the maximal n satisfying an >= 1/10.
Define ${a_n} = \left\lfloor {\sqrt {2n} + 0.5} \right\rfloor$ then $(a_n)^2$ is the denominator of the term of the sequence $T_n$.
However, the numerators are a bit trickery to find.

$T_{16}=\dfrac{1}{36}~T_{17}=\dfrac{3}{36}~\cdots~ T_{21}=\dfrac{11}{36}$

4. ## Re: sequence

...and the sum of each "group" = 1

5. ## Re: sequence

Write the elements of the sequence in the form of a lower triangular matrix (zeroes above the diagonal)

Row $n = \frac{2k-1}{n^2}$ where $1\leq k\leq n$ followed by a bunch of zeroes

The diagonal entry is the largest number in its row

so we need $\frac{2n-1}{n^2}<\frac{1}{10}$ which gives $n\geq 20$

The first 19 rows have a total of $1+2+...+19=190$ entries (not counting the zeroes)

Check: $\frac{37}{19^2}>\frac{1}{10}$ and $\frac{39}{20^2}<\frac{1}{10}$

6. ## Re: sequence

Nice "Idea"!

I made a silly goof in my post...
at numerator 15, started jumping by 2 instead of 1...
I'm standing in the corner!

Denominator 19^2 ends with 37/19^2 which is the 190th term.
So at 190th term, we have 37/19^2 = .1025....; still > 1/10.

The next bunch ends with 39/20^2 which is 210th term:
39/20^2 = .0975

So 190 is the man!

7. ## Re: sequence

Originally Posted by Idea
The diagonal entry is the largest number in its row
Or the largest number in a row is the last one;
like 9 / 5^2 for row#5.

(2u + 1) / [(2u +2)/2]^2 = 1 / 10
u = ~18.49
u = CEIL(u) = 19