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Let $f: \Bbb R \to \Bbb R $ be a monotonic decreasing function such that $\displaystyle{\lim_{x\to \infty}} f(x)= 0$. Prove that:

$$\lim_{n\to \infty} n \sum_{j=1}^n \frac{\cos\big(\frac{n}{j}\big)f\big(\frac{n}{j}\big)}{j^2} < \infty$$

My only thought about it was maybe using the integral test for convergence, but I didn't find a way to do it.

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    $\begingroup$ Hint: Riemann Sum. $\endgroup$
    – User8128
    Aug 21 '17 at 19:09
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Hint:

$$\frac{1}{n}\sum_{j=1}^{n}\frac{\cos\left(\frac{n}{j}\right)\,f\left(\frac{n}{j}\right)}{\left(\frac{j}{n}\right)^2}$$ is a Riemann sum associated with $$\int_{0}^{1}\frac{\cos\left(\frac{1}{x}\right)\,f\left(\frac{1}{x}\right)}{x^2}\,dx \stackrel{x\mapsto\frac{1}{z}}{=}\int_{1}^{+\infty}\cos(z)\,f(z)\,dz$$ which is convergent (in the improper Riemann integrability sense) due to Dirichlet's test: $f(x)$ is decreasing towards zero and $\cos(x)$ has a bounded primitive.

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  • $\begingroup$ I actually thought about Dirichlet's test but shouldn't $f$ be continuously differentiable in order to use it? $\endgroup$
    – user401516
    Aug 21 '17 at 19:13
  • $\begingroup$ @user401516: no, the monotonicity of $f$ is enough, see the last paragraph of the linked Wikipedia page. $\endgroup$ Aug 21 '17 at 19:15
  • $\begingroup$ And anyway, if $f$ is non-negative and decreasing towards zero, $f*K_\varepsilon$ has the same properties and it is continuously differentiable, it is enough to pick a smooth and concentrated kernel $K_\varepsilon$. $\endgroup$ Aug 21 '17 at 19:16
  • $\begingroup$ Strange, my professor proved this theorem when $f$ is continuously differentiable, but not non-negative though. $\endgroup$
    – user401516
    Aug 21 '17 at 19:31
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    $\begingroup$ @user401516: maybe because the proof is simpler in such a case, but the principle stays the same. The previous mollification argument shows that the $C^1$ assumption is not really needed. $\endgroup$ Aug 21 '17 at 19:34

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