Estimate the sum of a series only with the recursion $a_{n+1}=\frac{a_n}{a_n^2+1}$

Question

We have $\{a_n\}$,

$$ a_{1}=1,\quad a_{n+1}=\frac{a_n}{a_n^2+1}\quad \big(n\in\mathbf{N}^{*}\big) $$

Note

$$ S_n=\sum\limits_{k=1}^{n}a_k $$

How to estimate $S_n$? For example, I just want to know $\lfloor S_{2024}\rfloor$ without dirty number calculations. (But algebra calculation and inequality is welcome.)

So far I have known the upper bound could be estimated by

$$ \left(\frac{1}{a_{n}}\right)^2+2\le \left(\frac{1}{a_{n+1}}\right)^2$$

then

$$ a_n\le \frac{1}{\sqrt{2n}}\le \frac{\sqrt{2}}{\sqrt{n}+\sqrt{n-1}}\le \sqrt{2}\cdot (\sqrt{n}-\sqrt{n-1}) $$

to get

$$ S_n\le \sqrt{2n} $$

By the way, is there any website could calculate the sum of series only with recursion?

[ As a English beginner, maybe my expression should be corrected. :( ]

Answer 1

Not a complete solution, but here's a good start:

Since $a_{n+1} = \dfrac{a_n}{a_n^2+1}$, we have $\dfrac{1}{a_{n+1}} = \dfrac{a_n^2+1}{a_n} = \dfrac{1}{a_n}+a_n$, and thus, $\dfrac{1}{a_{n+1}}-\dfrac{1}{a_n} = a_n$ for all $n$.

Hence, $S_n = \displaystyle\sum_{k = 1}^{n}a_n = \sum_{k = 1}^{n}\left(\dfrac{1}{a_{k+1}}-\dfrac{1}{a_k}\right) = \dfrac{1}{a_{n+1}}-\dfrac{1}{a_1} = \dfrac{1}{a_{n+1}}-1$ for all $n$.

So, a good upper bound on the sequence $\{a_n\}$ will give us both upper and lower bounds on $S_n$.

As you noted, $\dfrac{1}{a_{n+1}^2} = \left(\dfrac{1}{a_n}+a_n\right)^2 = \dfrac{1}{a_n^2}+2+a_n^2 \ge \dfrac{1}{a_n^2}+2$ for all $n$.

Since $\dfrac{1}{a_2^2} = 4$, we can use induction to prove that $\dfrac{1}{a_n^2} \ge 2n$ for all $n \ge 2$.

Hence, $$\boxed{\sqrt{2n+2}-1 \le \dfrac{1}{a_{n+1}}-1 = S_n \le \sum_{k = 1}^{n}a_n \le 1+ \sum_{k = 2}^{n}\dfrac{1}{\sqrt{2k}}.}$$

For $n = 2024$, this gives $62.63 \le S_{2024} \le 62.90$, however, that involves evaluating the sum $\displaystyle\sum_{k = 2}^{2024}\dfrac{1}{\sqrt{2k}}$ exactly. If we use the bound $\displaystyle\sum_{k = 2}^{2024}\dfrac{1}{\sqrt{2k}} \le \displaystyle\sum_{k = 2}^{2024}(\sqrt{2k}-\sqrt{2k-2}) = \sqrt{4048}-\sqrt{2}$, then we get an upper bound of $S_{2024} \le 63.21$, which is not sharp enough to determine $\lfloor S_{2024} \rfloor$.

We can likely get sharper bounds if we bound $\displaystyle\sum_{k = 2}^{2024}\dfrac{1}{\sqrt{2k}} \le \sum_{k = 2}^{m}\dfrac{1}{\sqrt{2k}} + \sum_{k = m+1}^{2024}(\sqrt{2k}-\sqrt{2k-2}) = \sum_{k = 2}^{m}\dfrac{1}{\sqrt{2k}} + \sqrt{4048}-\sqrt{2m}$ for some reasonably small $m$. I think $m = 10$ will give us an upper bound less than $63$, but the calcuation is a bit messy.

Answer 2

Some thoughts.

We have $S_n = T_n - 1$ where $T_n$ is defined by: $T_1 = 2, T_n = T_{n-1} + \frac{1}{T_{n-1}}$ for $n\ge 2$.

We can use the approach in this (e.g. Did's answer) to estimate $T_n$. Also, see this.

Answer 3

Too long for a comment and just for your curiosity.

Starting from @JimmyK4542's nice answer, using generalized harmonic numbers $$S_n=\sum_{k = 2}^{n}\dfrac{1}{\sqrt{2k}}=\dfrac{1}{\sqrt{2}}\left(H_n^{\left(\frac{1}{2}\right)}-1\right)$$ Using asymptotics $$S_n=\sqrt{2n}+\frac{\zeta \left(\frac{1}{2}\right)-1}{\sqrt{2}}+\frac{1}{2 \sqrt{2n}}\left( 1-\frac{1}{12 n}+\frac{1}{192 n^3}+O\left(\frac{1}{n^5}\right) \right)$$ where $\zeta \left(\frac{1}{2}\right)=-1.46035$.

So, as @JimmyK4542 wrote, you do not need use many terms.