real-analysis
Let $f_n(x) =x^n$ for $n \in \mathbb{ N.}$ Which of the following statements are true?
$1)$ The sequence $\{f_n\}$ converges uniformly on $[ \frac{1}{6},\frac{1}{3}]$
$2)$ The sequence $\{f_n\}$ converges uniformly on $[ 0,1]$
$3)$ The sequence $\{f_n\}$ converges uniformly on $( 0,1)$
My answer is option 1) and option 3) because if $-1 <x <1$ ,then $|x|^n$ is converges to $0$ when n tend to infinity
Is it correct ??
I don't understand your attempt $f_n(x)=\begin{cases}1 &\text{ if }x\in (0,1)\\0 &\text{ if }x=0\end{cases}$. This equation is definitely wrong...
If you read carefully, then you see that you just have to decide which of $f_n$ and $g_n$ congerves uniformly.
First, you need to find the pointwise limit of $f_n$ and $g_n$. For each $x\in (0,1)$ you define $$ f(x):=\lim_{n\to\infty}f_n(x)\text{ and }g(x):=\lim_{n\to\infty}g_n(x). $$ This should be not so hard.
Next, you compute the difference of $f_n$ to $f$ and $g_n$ to $g$ with respect to the supremum norm: $$ \|f_n-f\|_\infty=\sup_{x\in(0,1)}|f_n(x)-f(x)|\text{ and }\|g_n-g\|_\infty=\sup_{x\in(0,1)}|g_n(x)-g(x)|. $$ Finally, you get $$ f_n\text{ converges uniformly to }f \Leftrightarrow \|f_n-f\|_\infty\to 0\\ \text{ and } g_n\text{ converges uniformly to }g \Leftrightarrow \|g_n-g\|_\infty\to 0. $$
User ModCon had a (now deleted) solution to some parts of the problem. There was a small mistake in it that made a few, but not all conclusions incorrect. I'll try to repeat his contributions, and add a few of my own.
We have $$\left|\frac{\sin(x/n)}n\right| \le \frac{|x/n|}n = \frac{|x|}{n^2}.$$
That means for each $x$ the series $\sum_{n=1}^{\infty}\frac{\sin(x/n)}{n}$ has a convergent majorant
$$\sum_{n=1}^{\infty}\frac{|x|}{n^2} = |x| \sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}6|x|,$$
which means it converges pointwise. This also means that we can use the Weierstrass M-Test on an interval $[-a,a]$ for some $a>0$ and choose $M_n=\frac{a}{n^2}$. That means the series converges absolutely and uniformly on $[a,-a]$, which means $f(x)$ is continuous on that inverval. Since we can chose any real $a$, this means $f(x)$ is continuous on the whole $\mathbb R$.
If we differentiate the series term-wise, we get another function:
$$g(x)=\sum_{n=1}^{\infty}\frac{\cos(x/n)}{n^2}$$
Using $|\cos(x/n)| \le 1$ we can directly apply the M-test on whole real line for $g(x)$, when setting $M_n=\frac1{n^2}$. This shows that g(x) is actually well defined and the defining series converges uniformly on $\mathbb R$ to $g(x)$.
So is now $f'(x)=g(x)$? By Theorem 1 on page 2 in this university script, the answer is yes. The term-wise derivate series (for $g(x)$) converges uniformly on the whole $\mathbb R$, the original series (for $f(x)$) converges on a point (we already know it converges anywhere), so we now know that
$$f'(x)=g(x),\; \forall x \in \mathbb R$$
This means $f(x)$ is differentible, which answers (c) in the affirmative.
We also know that
$$|g(x)| \le \sum_{n=1}^{\infty}\frac{|\cos(x/n)|}{n^2} \le \sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}6,$$
so the derivate of $f(x)$ is bounded. That means $f(x)$ is uniformly continuous: Given any $\epsilon > 0$, we can choose $\delta=\frac{6\epsilon}{\pi^2}$ and if we assume $x_0 < x_1 < x_0+\delta$, we have $f(x_1)=f(x_0)+(x_1-x_0)f'(\xi)$ with $\xi \in [x_0,x_1]$ (Mean Value Theorem), so
$$|f(x_1)-f(x_0)| = |f(x_0)+(x_1-x_0)f'(\xi) - f(x_0)| = |x_1-x_0||f'(\xi)| < \delta\frac{\pi^2}6 = \epsilon.$$
This ansers (b) in the affirmative and (a) in the negative.
For (d), because we know $f(x)$ is differentiable, it is enough to find a point where $f'(x)=g(x)$ is negative, to prove that (d) is not true.
Consider $x=\pi$. Then the first element of the series is $\cos(\pi)=-1$. We know that all the other elements of the series can sum up to at most $\sum_{n=\color{red}{2}}^{\infty}\frac{1}{n^2} = \frac{\pi^2}6 - 1 < 1$, so $g(\pi) < 0$ and (d) is not true.
If you don't want to use the non-elementary exact sum of that series, you can can also just say
$$\sum_{n=2}^{\infty}\frac{1}{n^2} < \sum_{n=2}^{\infty}\frac{1}{n(n-1)} = \sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n}\right) = (1-\frac12)+(\frac12-\frac13)+(\frac13-\frac14)\ldots=1.$$
Finally, a plot of the function (hopefully good enough truncated) by Wolfram Alpha:
Best Answer
The convergence is uniform only for $(1)$.
We have $$\sup_{x \in \left[\frac16, \frac13\right]}|x^n| = \frac1{3^n} \xrightarrow{n\to\infty} 0$$
so $f_n \to 0$ uniformly on $\left[\frac16, \frac13\right]$.
For $(2)$, the pointwise limit is $$f(x) = \begin{cases} 0, &\text{ if $x \in [0,1)$} \\ 1, &\text{ if $x = 0$}\end{cases}$$ which is not a continuous function. However, uniform limit of a sequence of continuous functions is always continuous.
For $(3)$ the pointwise limit is $0$ but for $x_n = 1- \frac1n$ we have
$$\sup_{x \in (0,1)} |x^n| \ge \left(1-\frac1n\right)^n \xrightarrow{n\to\infty} \frac1e$$
so $\sup_{x \in (0,1)} |x^n| \not\to 0$. Hence the convergence is not uniform.