Does this series converge uniformly?

[Lotto]

Homework Statement

Does series $\sum_{n=1}^{\infty} \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}}$ converges uniformly on the interval $[0,2\pi]$?

Relevant Equations

$\exists \epsilon >0 \, \forall n_0 \, \exists n \geq n_0 \, \exists p \, \exists x \in M: \left|\sum_{k=n+1}^{n+p} f_k(x)\right| \geq \epsilon$

First, I tried to show that $f_n$ converges uniformly on $[0,2\pi]$, which is true since $f_n \rightarrow 0$ for $n \rightarrow \infty$ and $\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0$. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that $\sin\left(\frac{n^2}{n+\frac 15}x \right)$ has partialy bounded sums on $[0,2\pi]$, which I think is not true. Weierstrass M-test doesn't obviously work.

So I tried to prove that $\exists \epsilon >0 \, \forall n_0 \, \exists n \geq n_0 \, \exists p \, \exists x \in [0,2\pi]: \left|\sum_{k=n+1}^{n+p} f_k(x)\right| \geq \epsilon$, because that would mean that my series doesn't converge uniformly. I tried to choose $n=n_0$, $p=n_0$ and $x=\frac{1}{n^2_0}$, but that didn't work.

Don't you have any tips how to solve this problem?

 

[anuttarasammyak]

FYI please share the plot of the sum as function of x which I ordered to ChatGPT.

 

[anuttarasammyak]

$x^2-3x+3>1$
$x<1,2<x$
for these area
$$ \sum_{n=1}^\infty |\sigma_n| < \sum_{n=1}^\infty \frac{1}{n^p} ,p>1$$
so the sum converges. Therefore we may pay our effort for $1 < x < 2$

PS

 

[Lotto]

$x^2-3x+3>1$
$x<1,2<x$
for these area
$$ \sum_{n=1}^\infty |\sigma_n| < \sum_{n=1}^\infty \frac{1}{n^p} ,p>1$$
so the sum converges. Therefore we may pay our effort for $1 < x < 2$

PS

View attachment 367033

Ah, I see, so on $[0,1)$ and $(2,2\pi]$ our series converges uniformly according to Weierstrass M-test. So for $[1,2]$ I would use Dirichlet's test. We know that $\frac{1}{n^{x^2-3x+3}}$ converges uniformly to zero on $[1,2]$. So if we show that $\sin\left(\frac{n^2}{n+\frac 15} x\right)$ has partially bounded sums, then according to Dirichlet's test the series converges uniformly on $[1,2]$ and also on $[0,2\pi]$.

But how do we show that $\sin\left(\frac{n^2}{n+\frac 15} x\right)$ has partially bounded sums? Using the complex definition of sine isn't good in this case, so how to do it?

 

[anuttarasammyak]

From the graph in ＃7, we observe for |x| >> 1 only n=1 term survives so it is sinusoidal between -1 and 1.
From the graph in #3 , for 1<x<2, S_100 shows vibration. It suggests a toughness of showing a convergence. I have no good idea to solve it.

 

[pasmith]

It may help to note that $$\frac{n^2}{n + \frac15} = n - \frac{1}{5 + \frac 1n}.$$