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Dec 7, 2021 at 16:47 comment added River Li Thanks. Your answer is helpful.
Feb 17, 2021 at 7:44 comment added juan @RiverLi I have not completed my arguments, I think what remains is easy. Read my entire entry to see how the initial assumptions may be proved.
Feb 17, 2021 at 7:43 comment added juan @RiverLi I do not assume that $x_n=n f(1/n)$, I assume $x_{2n}=nf(1/n)$ and $x:_{2n+1}=n g(1/n)$. My power series exist. If they have positive radius of convergence, and I think this is not very difficult to prove, then my assumptions about $x_{2n}$ and $x_{2n+1}$ are proved and all is justified.
Feb 16, 2021 at 23:46 comment added River Li Thanks for your work. But I am afraid that we may not assume that form of $f(x), g(x)$. 1st example: For the OP, if we assume $x_n = n + \frac{1}{2} + \sum_{k=1}^\infty \frac{a_k}{n^k}$, then we get $a_1 = \frac{5}{8}$ etc. without contradiction, but it is incorrect. 2nd example: For the recurrence relation $b_0 = 1, b_{n+1} = b_n + \frac{1}{b_n} + 2, \ n\ge 0$, we have $b_n \sim 2n + \frac{1}{2}\ln n + o(\frac{1}{n})$. Thus, there are other forms besides power series.
Feb 16, 2021 at 17:40 history answered juan CC BY-SA 4.0