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Let $a_1,a_2,a_3,…,a_n$ be a sequence of real numbers which satisfies the relation $a_{n+1}=\sqrt{a^2_n+1}$∀n∈N. Suppose that there exists a positive integer $n_0$ such that $a_{2{n_0}}=3a_{n_0}$. Then, find the value of $\lfloor \sum_{n=1}^{49} \sqrt{\frac{8}{8a^2_n +7}} \rfloor $ where the $\lfloor . \rfloor$ denotes the greatest integer function.

My attempt: For $n_0 =1$, we get the value of $a^2_1 = 1/8$.

Also using the relation, $a_n$ simplifies to $a^2_n = a^2_1 + n -1 =\frac {8n-7}{8}$.

Now substituting this value into the required sum, it reduces to $\lfloor \sum_{n=1}^{49} {\frac{1}{\sqrt n}} \rfloor $. Using Python I have verified that this is 12, which is the correct answer, however the question requires to do it manually, and I am unable to do so.

Any hints on how to approach solving further, or any other methods of going about the problem are appreciated.

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  • $\begingroup$ A rather unusual form of exercise... $\endgroup$
    – Jean Marie
    Commented Mar 21, 2021 at 17:44
  • $\begingroup$ Do you begin your sequence at index 1 (it looks so) or index 0 ? $\endgroup$
    – Jean Marie
    Commented Mar 21, 2021 at 17:49
  • $\begingroup$ @JeanMarie as per the question, I started at index 1 as it is defined for all natural numbers $\endgroup$
    – Hridai Khurana
    Commented Mar 21, 2021 at 17:51
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    $\begingroup$ @JeanMarie I do not think there is a value of $a_0$ which works on the $a_{2{n_0}}=3a_{n_0}$ condition $\endgroup$
    – Henry
    Commented Mar 21, 2021 at 17:52
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    $\begingroup$ @JeanMarie: there is no clear consensus as to whether $\mathbb{N}$ contains $0$. I myself prefer $0\not\in\mathbb{N}$, but it is good when it is stated what the assumption is. $\endgroup$
    – robjohn ♦
    Commented Mar 21, 2021 at 17:59

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Hints:

  • Consider $b_n=a_n^2$ and rewrite the question in terms of $b_n$
  • You need to find which values of $n_0$ work and give a credible value of $a_1$. It look to me as if $n_0=1$ does.

Added:

  • You have since found $a_n=\frac{n}{8}-\frac{7}{8}$. This makes $\lfloor \sum_{n=1}^{49} \sqrt{\frac{8}{8a^2_n +7}} \rfloor = \sqrt{1}+\sqrt{\frac12}+\sqrt{\frac13}+ \cdots+\sqrt{\frac{1}{48}}$ rounded down which is not easy by hand, but $\int\limits_{1}^{49} \frac 1 {\sqrt{x}}\,dx = 12$ and $\int\limits_{0.25}^{49} \frac 1 {\sqrt{x}}\,dx = 13$ so the answer is going to be $12$ when rounded down, using $\int\limits_{n}^{n+1} \frac 1 {\sqrt{x}}\,dx < \frac1{\sqrt{n}} < \int\limits_{n-0.75}^{n+0.25} \frac 1 {\sqrt{x}}\,dx$
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  • $\begingroup$ Thanks. Using these I've reduced the sum, I'm new to stack exchange, so should I edit my question to show the progress or include it in a comment? $\endgroup$
    – Hridai Khurana
    Commented Mar 22, 2021 at 5:12
  • $\begingroup$ @HridaiKhurana Editing your question for progress would be good - if you had a full answer, you could answer your own question $\endgroup$
    – Henry
    Commented Mar 22, 2021 at 9:38
  • $\begingroup$ I've made the edit, hope its clear $\endgroup$
    – Hridai Khurana
    Commented Mar 22, 2021 at 11:24
  • $\begingroup$ @HridaiKhurana I have suggested a way of finding the the answer is between $12$ and $13$ so $12$ when rounded down $\endgroup$
    – Henry
    Commented Mar 22, 2021 at 11:48
  • $\begingroup$ Thanks! I've accepted the answer. However, is there a reason to why you took the limit to be -0.75 of the lower bound, or is it a theorem I am unaware of? $\endgroup$
    – Hridai Khurana
    Commented Mar 22, 2021 at 12:09

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