Does $\lim \limits_{n \rightarrow \infty} n \sum \limits_{j=1}^{n} \frac{\cos(\frac{n}{j})f(\frac{n}{j})}{j^2}$, $f\in C^1$, exist?

Question

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 $$\lim \limits_{n \rightarrow \infty} n \sum \limits_{j=1}^{n} \frac{\cos(\frac{n}{j})f(\frac{n}{j})}{j^2}$$ 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. 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 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 $\cos(\frac{1}{x})f(\frac{1}{x})x^{-2}$ was Riemann-integrable on $[0,1]$.

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?

The key here is that you have nowhere used the assumption that $f$ was monotonically decreasing with limit $0$. That should help... — Clement C., Jul 30 '18 at 23:15
But I have, the Dirichlet convergence test requires f to be monotonically decreasing — Yuval Gamzon, Jul 31 '18 at 04:37
Indeed, taking $f(x)=1/x$, we have $S_n = \sum_{j=1}^{n} \cos(n/j)/j,$ which shouldn't converge by comparison to the harmonic series. — Integrand, Nov 04 '24 at 17:13
Yes but $f(x)=1/x$ is not $\mathcal{C}^1$ on whole $\mathbb{R}$ — EDX, Nov 04 '24 at 18:59

EDX · Answer 1 · 2024-11-04T19:03:56.883

For Riemann-sum to improper check this out https://math.stackexchange.com/a/1745876/763728.

Once ok, following the current question undertaken :

$$ \int_{1}^{\infty} f(x)\cos(x)dx < \infty $$

is equivalent to

$$ \int_{\pi/2}^{\infty} f(x)\cos(x)dx < \infty $$

Writting so :

$$ \int_{\pi/2}^{\infty}f(x)\cos(x)dx= \sum_{n=1}^{\infty}(-1)^n\int_{(n-1/2)\pi}^{(n+1/2)\pi}f(x)|\cos(x)|dx$$

Theorem of convergence of alternate series :

Theorem (Alternated series) Let a sequence terms of a serie be : $$u_n=(-1)^na_n $$ Such that $a_n$ is decreasing and tending to $0$. Then$$ \sum_{n=0}^{\infty}u_n$$ converges.

So define :

$$ a_n \triangleq \int_{(n-1/2)\pi}^{(n+1/2)\pi}f(x)|\cos(x)|dx>0$$ (because $f$ monotically decreasing to $0$)

then we have straightforwardly :

$$ a_{n+1} \leq a_{n} $$

And as well :

$$ a_n \leq \pi \max\left(f((n-1/2)\pi),f((n+1/2)\pi)\right) \to 0$$

Hence

For a continously differentiable function $f\in \mathcal{C}^1$ :

$$ \int_{1}^{\infty}f(x)\cos(x)dx < \infty $$

The limit of the Riemann sum presented above.

Does $\lim \limits_{n \rightarrow \infty} n \sum \limits_{j=1}^{n} \frac{\cos(\frac{n}{j})f(\frac{n}{j})}{j^2}$, $f\in C^1$, exist?

1 Answers1