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  1. 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 ?

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rtybase
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Aster Zen
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    This is not clear at all. You appear to have defined a sequence ${a_n}$ for some $n$ though not all. There's no apparent reason why the definition shouldn't go on forever. None of this appears to have anything to do with $n≤\frac 1{10}$. – lulu Nov 24 '18 at 14:35
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    Voting to close the question as it is unclear what you are asking. If you can, please edit for clarity. – lulu Nov 24 '18 at 14:51
  • @lulu I have taken a picture to give it more clear but I do not know how to attach this picture, can u help me ? I am a newbie, thanks before – Aster Zen Nov 24 '18 at 14:55
  • I don't know how to attach pictures, sorry. Never done that. – lulu Nov 24 '18 at 14:57
  • If you're using a browser, there's an icon at the top left of the edit field that looks like two mountains, one bigger than the other - the standard icon for picture, these days, infact - click on that & follow instructions. As for your series, it seems that values both arbitrarily small and arbitrarily close to 1/2 (but<1/2) occur infinitely in it. – AmbretteOrrisey Nov 24 '18 at 16:59
  • @AmbretteOrrisey ok thanks I have given my picture now, thank u so much – Aster Zen Nov 25 '18 at 00:18
  • @lulu, I have fixed my question, can you help me? the question is asked for n >= by the way – Aster Zen Dec 16 '18 at 08:25

2 Answers2

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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 $$\boxed{a_{190}=\frac{37}{361}}$$

bof
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  • The short answer is $n=190$ ;-) –  Dec 16 '18 at 09:49
  • Thankyou @bof, you explained it very clear ! But I wonder about the sequence, can we neglect the 1/4, 1/9, 3/9 , ? Actually I have found that on the same denominator if I sum all of it the value is one – Aster Zen Dec 16 '18 at 14:43
  • @AsterZen Since you're looking for the last term in the sequence which is greater than or equal to 1/10, we can disregard the 1/4, 1/9, 3/9 because each of them is immediately followed by a bigger term. So we just look at the last and biggest term with each denominator. – bof Dec 17 '18 at 00:10
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The general terms are

$$\frac{2k+1}{m^2}$$ for $k\in[0,m-1]$, and the largest of a subsequence is

$$\frac{2m-1}{m^2}=\frac2m-\frac1{m^2}.$$

By inspection, $m=20$ is just too large ($\dfrac1{10}-\dfrac1{400}$) so that $19$ is fine. The corresponding $n$ is the $19^{th}$ triangular number, $\color{green}{190}$.

Just for fun, the general expression can be written

$$\frac{2n-\left\lceil\dfrac{\sqrt{8n+1}-1}2\right\rceil\left(\left\lceil\dfrac{\sqrt{8n+1}-1}2\right\rceil-1\right)-1}{\left\lceil\dfrac{\sqrt{8n+1}-1}2\right\rceil^2}.$$