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 $$\frac{n^2}{n + \frac15} = n - \frac{1}{5 + \frac 1n}.$$

 

[anuttarasammyak]

It may help to note that

As for the special case $x=\frac{\pi}{2}$

for n=0 (mod 4)
$$a_n=-\frac{1}{n^b}\sin c_n$$

for n=1 (mod 4)
$$a_n=\frac{1}{n^b}\cos c_n$$

for n=2 (mod 4)
$$a_n=\frac{1}{n^b}\sin c_n$$

for n=3 (mod 4)
$$a_n=-\frac{1}{n^b}\cos c_n$$

where
$$c_n=\frac{1}{5+\frac{1}{n}}\frac{\pi}{2} = \frac{\pi}{10}(1-\frac{1}{5n}+o(n^{-2}))$$
$$b=(\frac{\pi}
{2}-\frac{3}{2})^2+\frac{3}{4}=0.755501...$$

The partial sum for Dirichlet's test is
$$S_N=\cos c_1 + \sin c_2 - \cos c_3 - \sin c_4 + ...$$
$$ = 2\sin \frac{c_1+c_3}{2}\sin\frac{c_3-c_1}{2}-2\cos\frac{c_2+c_4}{2}\sin\frac{c_4-c_2}{2}+...$$
This seems bounded but should be investigated,

 

[julian]

To show that $\sin (\frac{n^2}{n+a} x)$ has partially bounded sums for $x \in [1,2]$: Write

\begin{align*}
\frac{n^2}{n+a} = n - a + \frac{a^2}{n+a}
\end{align*}


Then

\begin{align*}
\sin (\frac{n^2}{n+a} x) = \sin (n - a + r_n) x , \quad where \quad r_n = \frac{a^2}{n+a}
\end{align*}

Complex exponentials:


\begin{align*}
S_N = \sum_{n=1}^N e^{i (n - a + r_n) x} = e^{-ia x } \sum_{n=1}^N e^{inx} e^{i r_n x}
\end{align*}


Write


\begin{align*}
\sum_{n=1}^N e^{inx} e^{i r_n x} = \sum_{n=1}^N e^{inx} + \sum_{n=1}^N e^{inx} (e^{i r_n x} - 1)
\end{align*}


$\sum_{n=1}^N e^{inx}$ is a partialy bounded sum. Then consider the sum:


\begin{align*}
\sum_{n=1}^N e^{inx} (e^{i r_n x} - 1) & = \sum_{n=1}^N e^{inx} i r_n x + \sum_{n=1}^N e^{inx} \sum_{m=2}^\infty \frac{(ix)^m}{m!} \frac{a^{2m}}{(n+a)^m}
\nonumber \\
& = \sum_{n=1}^N e^{inx} i r_n x + \sum_{m=2}^\infty \frac{a^{2m}}{m!} \sum_{n=1}^N e^{inx} \frac{(ix)^m}{(n+a)^m}
\end{align*}


First sum: $r_{n+1} x \leq r_n x$ for all $x \in [1,2]$. By Dirichlet convergences and hence bounded partial sums. For the second sum:

\begin{align*}
|\sum_{m=2}^\infty \frac{a^{2m}}{m!} \sum_{n=1}^N e^{inx} \frac{(ix)^m}{(n+a)^m}|
& \leq \sum_{m=2}^\infty \frac{x^m a^{2m}}{m!} \sum_{n=1}^N \frac{1}{(n+a)^m}
\nonumber \\
& \leq \sum_{m=2}^\infty \frac{x^m a^{2m}}{m!} \sum_{n=1}^\infty \frac{1}{(n+a)^m}
\nonumber \\
& \leq \sum_{m=2}^\infty \frac{x^m a^{2m}}{m!} \sum_{n=1}^\infty \frac{1}{n^m}
\end{align*}

But $\lim_{m \rightarrow \infty} \zeta(m) = 1$, so by the limit comparison test we have convergence.