I stumbled upon this question from a Calculus exam: Let $f \in C^1(\mathbb{R})$ be monotonically decreasing such that $\lim \limits_{x \rightarrow \infty} f(x) = 0$. Prove that the limit $$\lim \limits_{n \rightarrow \infty} n \sum \limits_{j=1}^{n} \frac{\cos(\frac{n}{j})f(\frac{n}{j})}{j^2}$$ exists and is finite.

Now obviously the solution the writers of this problem had intended was to look at the integral $$\int \limits_{0}^{1} \frac{\cos(\frac{1}{x})f(\frac{1}{x})}{x^2}dx = \int \limits_{1}^{\infty} f(x)\cos x \, dx$$ which converges by the Dirichlet criterion and then the Riemann sums attributed to the partitions $\Pi_n = \{0, \frac{1}{n}, \frac{2}{n}, ..., 1 \}$ are $S_n=n \sum \limits_{j=1}^{n} \frac{\cos(\frac{n}{j})f(\frac{n}{j})}{j^2}$ which if would imply $$\lim \limits_{n \rightarrow \infty} n \sum \limits_{j=1}^{n} \frac{\cos(\frac{n}{j})f(\frac{n}{j})}{j^2}=\int \limits_{0}^{1} \frac{\cos(\frac{1}{x})f(\frac{1}{x})}{x^2}dx < \infty$$ if the function $\frac{\cos(\frac{1}{x})f(\frac{1}{x})}{x^2}$ was Riemann-integrable in $[0,1]$.

The problem with that approach however, is that the integral $$\int \limits_{0}^{1}\frac{cos(\frac{1}{x})f(\frac{1}{x})}{x^2}dx$$ is not a Riemann-integral but an improper-integral so one cannot conclude that $$\lim \limits_{n \rightarrow \infty} S_n = \int \limits_{0}^{1}\frac{\cos(\frac{1}{x})f(\frac{1}{x})}{x^2}dx$$

After many hours of thinking, trying to resolve this issue I realized that the question may be false as it makes no sense that the limit exists because that would sort of mean that the integral is a proper Riemann-integral which would imply the function is bounded (which is not necessarily true depending on the choice of $f$). To test this, I used Wolfram Alpha to calculate numerical estimates with the function $f(x)=\frac{1}{x}$, which confirmed my speculations, though I have not been able to rigorously prove it.

My question is can you find an example that disproves the claim the question makes? Or was I wrong and the claim can be proven?

  • $\begingroup$ The key here is that you have nowhere used the assumption that $f$ was monotonically decreasing with limit $0$. That should help... $\endgroup$ – Clement C. Jul 30 '18 at 23:15
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    $\begingroup$ But I have, the Dirichlet convergence test requires f to be monotonically decreasing $\endgroup$ – Yuval Gamzon Jul 31 '18 at 4:37
  • $\begingroup$ Good point, I had missed that. $\endgroup$ – Clement C. Jul 31 '18 at 4:40

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