Find the maximal $n$ satisfying $a_n \geq \frac{1}{10}$ 

  
*Let $a_n$ be the $n$-th term of the following sequence $$\frac{1}{1},\frac{1}{4},\frac{3}{4},\frac{1}{9},\frac{3}{9},\frac{5}{9},\frac{1}{16},\frac{3}{16},\frac{5}{16},\frac{7}{16},\frac{1}{25},...$$

From what could I start resolving this problem ? I have found the sequence pattern that the numerator always the odd number which always start again from number 1 if the denominator change pattern. the pattern of the denominator is the square of 1, 2, 3 and etc
I have also found the pattern for the series that


*

*$a_1 = 1$

*$a_3= 2$

*$a_6 = 3$

*$a_{10} = 4$

*$a_{15}= 5$

*$a_{21} = 6$
And etc.
But I stucked on this formula, I have no idea to find the maximum n, can anyone give me some suggestion and steps for solving this problem ?


The terms of your sequence which are immediately followed by a smaller term are
$$\frac11,\ \frac34,\ \frac59,\ \frac7{16},\ \frac9{25}\ \dots$$
The $k^\text{th}$ one of these is given by the formula $\frac{2k-1}{k^2}$. The last term in your sequence which exceeds or equals $\frac1{10}$ will correspond to the greatest value of $k$ satisfying the inequality $\frac{2k-1}{k^2}\ge\frac1{10}$ or equivalently
$$f(k)=k^2-20k+10\le0$$
Solving the quadratic equation, the zeros of $f(k)$ are $10\pm\sqrt{90}$, so $f(k)\le0$ for $10-\sqrt{90}\le k\le10+\sqrt{90}$. Since the greatest integer below $10+\sqrt{90}$ is $19$, the last term of your sequence above $\frac1{10}$ is
$$\frac{2\cdot19-1}{19^2}=\frac{37}{361}$$
and the position of this term in your sequence is $1+2+3+\cdots+19=190$,
so the short answer to your question is
$$\boxed{a_{190}=\frac{37}{361}}$$
