Comparing Derivatives

The strategy for this problem is write $f(\frac{1}{n})$ in terms of the Taylor expansion of $f$ about $x = 0$ and then rearrange terms until we get something that allows us to make a conclusion about the odd derivatives of $f$ . First, note that since $f$ is continuous, $f(0) = \lim_{n \to \infty} f(\frac{1}{n}) = \lim_{n \to \infty} \frac{n^2}{n^2+1} = 1$ . $\begin{aligned} f(\frac{1}{n}) = \frac{n^2}{n^2 + 1} &= f(0) + \frac{f'(0)}{1!n} + \frac{f^{(2)}(0)}{2!n^2} + \frac{f^{(3)}(0)}{3!n^3} + \cdots \\ \frac{n^2}{n^2 + 1} &= 1 + \frac{f'(0)}{1!n} + \frac{f^{(2)}(0)}{2!n^2} + \frac{f^{(3)}(0)}{3!n^3} + \cdots \\ \frac{-1}{n^2 + 1} &= \frac{f'(0)}{1!n} + \frac{f^{(2)}(0)}{2!n^2} + \frac{f^{(3)}(0)}{3!n^3} + \cdots \\ -1 &= \frac{f'(0)}{1!n}(n^2 + 1) + \frac{f^{(2)}(0)}{2!n^2}(n^2 + 1) + \frac{f^{(3)}(0)}{3!n^3}(n^2 + 1) + \cdots \\ -1 &= \frac{f'(0)}{1!}n + \frac{f^{(2)}(0)}{2!} + \left( \frac{f'(0)}{1!} + \frac{f^{(3)}(0)}{3!} \right) \frac{1}{n} + \left( \frac{f^{(2)}(0)}{2!} + \frac{f^{(4)}(0)}{4!} \right) \frac{1}{n^2} + \left( \frac{f^{(3)}(0)}{3!} + \frac{f^{(5)}(0)}{5!} \right) \frac{1}{n^3} + \cdots \end{aligned}$

The coefficients of all the odd powers of $n$ are zero, so $f'(0) = 0$ , which implies $f^{(3)}(0) = 0$ which, in turn, implies $f^{(5)}(0) = 0$ , and so on. Note that you can likewise determine the even derivatives of $f$ , but since $f''(0) \neq 0$ , instead of all being zero, the will form an alternating sequence.