IMC 1997 · Day 2 · P9

Set $f(t) = \dfrac{\sin{(\log t)}}{t^\alpha}$. We have $$f'(t) = -\frac{\alpha}{t^{\alpha+1}} \sin{(\log t)} + \frac{\cos{(\log t)}}{t^{\alpha+1}}.$$ So $|f'(t)| \le \dfrac{1+\alpha}{t^{\alpha+1}}$ for $\alpha > 0$. Then from Mean value theorem for some $\theta \in (0,1)$ we get $|f(n+1) - f(n)| = |f'(n+\theta)| \le \dfrac{1+\alpha}{n^{\alpha+1}}$. Since $\sum \dfrac{1+\alpha}{n^{\alpha+1}} < +\infty$ for $\alpha > 0$ and $f(n) \xrightarrow[n \to \infty]{} 0$ we get that $\sum\limits_{n=1}^{\infty} (-1)^{n-1} f(n) = \sum\limits_{n=1}^{\infty} (f(2n-1) - f(2n))$ converges.

Now we have to prove that $\dfrac{\sin{(\log n)}}{n^\alpha}$ does not converge to 0 for $\alpha \le 0$. It suffices to consider $\alpha = 0$.

We show that $a_n = \sin{(\log n)}$ does not tend to zero. Assume the contrary. There exist $k_n \in \mathbb{N}$ and $\lambda_n \in \left[ -\frac{1}{2}, \frac{1}{2} \right]$ for $n > e^2$ such that $\dfrac{\log n}{\pi} = k_n + \lambda_n$. Then $|a_n| = \sin \pi |\lambda_n|$. Since $a_n \to 0$ we get $\lambda_n \to 0$. We have $$\begin{align*} k_{n+1} - k_n &= \frac{\log(n+1) - \log n}{\pi} - (\lambda_{n+1} - \lambda_n) \\ &= \frac{1}{\pi} \log \left( 1 + \frac{1}{n} \right) - (\lambda_{n+1} - \lambda_n). \end{align*}$$ Then $|k_{n+1} - k_n| < 1$ for all $n$ big enough. Hence there exists $n_0$ so that $k_n = k_{n_0}$ for $n > n_0$. So $\dfrac{\log n}{\pi} = k_{n_0} + \lambda_n$ for $n > n_0$. Since $\lambda_n \to 0$ we get contradiction with $\log n \to \infty$.