Help on a Putnam Problem from the 90s

Lemma: If $g\in C^\infty(\mathbb R),$ $a_1 > a_2 > \cdots \to 0,$ and $g(a_n) = 0$ for all $n,$ then $g^{(k)}(0) = 0$ for all $k.$

Proof: We have $g(0)=0$ by continuity. By Rolle's theorem, for each $n,$ $g'(c_n) = 0$ for some $c_n\in (a_{n+1},a_n).$ Note that $c_1>c_2>\ \cdots \to 0.$ This implies $g'(0)=0$ by continuity of $g'.$ Now $g'\in C^\infty$ and vanishes along the sequence $c_n,$ so we can apply the same argument to it to conclude $g''(d_n) = 0, d_n\in (c_{n+1},c_n).$ We again get $d_1>d_2 > \cdots \to 0,$ so $g''(0)=0.$ This process can be repeated indefinitely, giving the conclusion.

Back to our problem: Let $g(x) = f(x)-1/(1+x^2).$ Then $g\in C^\infty$ and $g(1/n) = 0$ for all $n.$ From the lemma, $g^{(k)}(0) = 0$ for all $k.$ Thus the derivatives of $f$ at $0$ equal the corresponding derivatives of $1/(1+x^2).$ Since $1/(1+x^2) = 1-x^2 + x^4 - \cdots$ for $|x|<1,$ the derivatives of $f$ can be read off from the power series as $f^{(2k)}(0) = (-1)^k(2k)!, f^{(2k+1)}(0)=0$ for all $k.$

If $f(x)$ is analytic (around $0$), we would then have $f(x)=\frac1{1+x^2}$

See then the geometric series expansion of $f(x)$.

$$f(x)=\frac1{1+x^2}=\sum_{n=0}^\infty(-1)^nx^{2n}$$

By Taylor's theorem,

$$f^{(k)}(0)=\begin{cases}0;&k\text{ is odd}\\k!;&(k\mod4)=0\\-(k!);&{(k\mod4)=2}\end{cases}$$

Partial Answer, way too long for a comment:

$f'(0)$ is easy to calculate:

By the mean value Theorem, there exists $c_n \in (\frac{1}{2n}, \frac{1}{n})$ such that

$$f'(c_n)=\frac{f(\frac{1}{n})-f(\frac{1}{2n})}{\frac{1}{n}-\frac{1}{2n}}=2n \left(\frac{n^2}{n^2+1}-\frac{4n^2}{4n^2+1}\right)=-\frac{6n^3}{(n^2+1)(4n^2+1)}$$

Now, since $f$ is twice differentiable, $f'$ is continuous at $x=0$.

By the squeeze Theorem, $\lim_n c_n=0$.

Therefore, as $f'$ is continuous at $x=0$, we have

$$f'(0)=\lim_n f'(c_n) =0$$

For $f''(0)$ Use the following version of MVT:

If $f$ is twice differentiable on $[a,b]$ and $f'$ is continuous on $(a,b)$, then there exists some $c \in (a,b)$ such that $$ f(b)-2 f\left(\frac{a+b}{2}\right)+f(a)=\frac{1}{4} (b-a)^2 f''(c)$$

Now, the tricky part is to pick $a=\frac{1}{n}$ and $b=\frac{1}{m}$ such that $\frac{a+b}{2}=\frac{1}{k}$.

One way to do this is to start from $$\frac{1}{3}+\frac{1}{6}=\frac{1}{2} \Rightarrow \\ \frac{1}{3n}+\frac{1}{6n}=\frac{1}{2n} =\frac{2}{4n} \Rightarrow \\ \frac{\frac{1}{3n}+\frac{1}{6n}}{2} =\frac{1}{4n} $$

Use the generalized version of MVT for $a=\frac{1}{6n}$ and $b=\frac{1}{3n}$ to get a formula for $f''(c_n)$, and then use that by the same argument as above we have $$f''(0)=\lim_n f''(c_n)$$

Unfortunately, this method cannot most likely be extended too much,but who knows. For each order of derivatives there should be a version of MVT, but it will most likely be tricky to make sure that all terms are of the given form. And the most likely replacement for the generalized MVT, the Lagrange remainder estimate for the Taylor remainder, includes derivatives at $\frac{1}{n}$ in the formula, which you cannot most likely not get. (you could apply this formula at $x=\frac{1}{n}$ or $x=0$, but each leads to the same issue).

But maybe someone can get an idea from here and find a solution.