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Let $$ \lim_{n\to\infty} n\cdot \sum_{j=1}^n \frac{\cos\left(\frac{n}{j}\right)f\left(\frac{n}{j}\right)}{j^2} $$

Where $f$ is $C^\infty$ and monotonically decreasing: $\lim_{x\to\infty} f(x) = 0$.

I need to evaluate the limit. Riemann integral should be used. I guess there should be some algebraic moves to reach that, and the integrand (my guess) should be $f(x)\cos(x)$.

Could you help me please connect the dots?

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Define $g(x)=\frac{\cos\left(\frac{1}{x}\right)f\left(\frac{1}{x}\right)}{x^2}$, and write the above as:

$$\sum_{j=1}^{n} \frac{1}{n}g\left(\frac{j}{n}\right)$$

Which is a Reimann sum, which has, as a limit:

$$\int_{0}^1 g(x)\,dx$$

That's gonna depend on $g$. In particular, it might be an improper integral, depending on whether you can make $g$ continuous at $0$.

Then $g\left(\frac{1}u\right)=u^2f(u)\cos u$

Substituting $x=\frac{1}{u}$ so $dx=\frac{-du}{u^2}$ you get: $$\int_1^{\infty} f(u)\cos u \,du$$

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  • $\begingroup$ I get: $$\sum_{j=1}^n \frac{\cos(\frac{n}{j})f(\frac{n}{j})}{\frac{j}{n^2}} = n\cdot \sum_{j=1}^n \frac{\cos(\frac{n}{j})f(\frac{n}{j})}{\frac{j}{n}}$$ $\endgroup$
    – AlonAlon
    Jan 16 '15 at 20:05
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    $\begingroup$ Whoops, missed the square. Just a second. $\endgroup$ Jan 16 '15 at 20:06
  • $\begingroup$ Could you also comment about your intuition about how to figure out what's the integral is? $\endgroup$
    – AlonAlon
    Jan 16 '15 at 20:07
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    $\begingroup$ As a general rule, when I see a lot of $j/n$ in a sum, I try to replace them by $x$ to get a function for which it is a Reimann sum. Usually a guess. $\endgroup$ Jan 16 '15 at 20:09
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    $\begingroup$ The answer should be fixed. @AlonAlon $\endgroup$ Jan 16 '15 at 20:10

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