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This is what I understand from @KaviRamaMurthy 's comment. Since $\sum_{n=0}^\infty c_nx^n$ is the real Taylor series for the integral $$\int_1^\infty t^{x-1}e^{-t}dt$$ and this integral converges for all positive $x$, so the Taylor series also converges for all positive $x$. Now we choose any $z\in\mathbb C$ and realize that $$\sum_{n=0}^\infty |c_nz^n|\leq ...


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Since a.s.-convergence implies convergence in probability, we can choose $m_k$ such that $$\mathbb{P}(\left|X_{m_k} - X\right|>2^{-k-1})<2^{-k-1}.$$ Similarly, we can choose $n_k$'s such that $$\mathbb{P}(\left|X_{m_k}^{n_k} - X_{m_k}\right|>2^{-k-1})<2^{-k-1}.$$ Then by the triangle inequality, \begin{align*} \mathbb{P}(\left|X_{m_k}^{n_k} - X\...


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First note that $\delta_{nk}$ has nothing to do with $k$, so write $\delta_n$ instead. Note also that, by construction, $\delta_n\downarrow 0$. Fix an irrational $\alpha$ and let $\epsilon>0$. There are only finitely many $n$ with $\frac1n>\epsilon$, so choose some $n_1$ such that $\frac1{n_1}<\epsilon$. If $\alpha$ is distance $>\delta_{n_1}$ ...


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Suppose there is some irrational $x$ such that $f_n(x)\not\to f(x)=0$. That is, for every $\epsilon<0$ and every $N\in \mathbb N$ we can find some positive integer $m>N$ such that $f_m(x)\geq \epsilon$. Thus there is some subsequence $\{f_{n_i} \}$ such that $f_{n_i}\geq \varepsilon$ for some positive real $\varepsilon$. By the construction of the $f_{...


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a) First, we show pointwise convergence. For any $x$ we build increasing $x_k \to x$ and decreasing $x'_k \to x$, belonging to $A$. For any $n$, we have $$f_n(x_k) \le f_n(x) \le f_n(x'_k)$$ Since $f_n$ converges on $A$, $$\forall k\ \forall \epsilon\ \exists n_{k,\epsilon}:\ \forall n > n_{k,\epsilon} \ \ |f_n(x_k) - f(x_k)| < \frac \epsilon 2 \tag{1}...


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Let us assume that $T_n$ does not converge uniformly on every compact set. Then there exists a compact set $K$, an $ε_0 >0$ and a subsequence $(T_n)_{n \in M \subset \mathbb N}$ of the original sequence, as well as a sequence $(x_n)_{n \in M} \subset K$ such that $|| T_n (x_n) -L(x_n)|| \geq ε_0$, for all $n \in M$. Since $K$ is compact, we may assume ...


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Uniformity is not needed, but while there is a difference between almost uniform convergence and uniform convergence almost everywhere, there is no material reason to introduce the terminology almost pointwise convergence: a sequence $\{f_n\}_{n\in\Bbb N}$ such that for all $\delta>0$ there is some $A$ such that $\mu(X\setminus A)<\delta$ and $\{\left....


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Most easy is to use sufficient and necessary condition of uniform convergence $$\lim_\limits{n \to \infty}\sup_{x \in \mathbb{R}}|f_n(x)-f(x)|=0$$ In your example function $s_n(x)=\frac{2}{3+n|x|}$ is even on $\mathbb{R}$, and you can find, that sup reached in $x=0$. Hence?


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Note that we have $$\begin{align} \int_0^1 \sum_{n=0}^N (-1)^n e^{-nx}\,dx&=\int_0^1 \frac{1+(-1)^N e^{-(N+1)x}}{1+e^{-x}}\,dx \end{align}$$ We remark that the function $\displaystyle f_N(x)=\frac{1+(-1)^N e^{-(N+1)x}}{1+e^{-x}}$ converges pointwise Almost Everywhere to $\displaystyle \frac1{1+e^{-x}}$ for $x\in [0,1]$. And inasmuch as $\displaystyle \...


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Indeed your made a very nice observation that is often neglected by practitioners of arcane art of integral and series. This type of technical issue is often overcome by realizing the given expression as the limit of perturbed expressions with additional parameters. (In this regard, we might possibly borrow the physics jargon 'regularization' for this ...


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1. Take $$f_n(x) = \frac{1}{1+n(1-x)}.$$ Then $f_n(x) \downarrow 0$ if $x\in[0,1)$ and $f_n(1)=1$. The following figure demonstrates the graph of $f_n$ on $[0,1]$ for $n=2^1,2^2,\dots,2^{14}$, colored in increasing order of frequency. 2. Take $$f_n(x) = \frac{2n}{1+(n(x-1)+\sqrt{2n-1})^2}.$$ Then $f_n(x) \to 0$ if $x \neq 1$ and $f_n(1)=1$. This example is ...


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I'll do you one better and work with a sequence with a pointwise limit that makes sense on all of $\mathbb{R}$. Consider the sequence of compactly supported smooth functions defined by $$ f_n(x) = \exp(1)\cdot\exp\left(-\frac{1}{1-(n(1-x))^2}\right).$$ on $\left(1-\frac{1}{n},1+\frac{1}{n}\right)$ and $0$ elsewhere.


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If you have $f_n(t)=t^n$,t$\in$[0,1] and you take a continuous g:[0,1]->$\mathbb R$ with g(1)=0(every function g with those conditions)then you can show that for every t$\in$[0,1] $(f_n.g)(t)$-->0 uniformly


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Let $g(x)$ be any function such that $g(x)\in[0,1)$ for $x\in[0,1)$ and $g(1)=1$. For example, $$g(x)=x^k$$ $$g(x)=sin\left({\pi x \over 2}\right)^k$$ $$g(x)={x^2+1\over 2}$$ $$g(x)=3x^2-2x^3$$ $$g(x)=6x^5-15x^4+10x^3$$ etc. Then, the sequence $$f_n(x) = g(x)^n$$ satisfies your requirements.


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Take $f_n(x) = (x^a)^n $ on $[0,1]$ , where $ a $ is any positive integer. Take $f_n(x) =( \sin(\frac{πx}{2}) )^n $ on $[0,1]$ Also, take $f_n(x) = (e^{x-1})^n $ on $[0,1]$ Take $f_n(x) =( \tan(\frac{πx}{4}) )^n $ on $[0,1]$


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hint $ f_n $ is well defined for $ n>0$ and for $ x\in [0,1] $ But, $$f_n(0)=n, $$ $$\lim_{n\to+\infty}f_n(0)=+\infty$$ and for $ x\in (0,1],$ $$\lim_{n\to+\infty}f_n(x)=f(x)$$ So, the convergence of $ (f_n) $ to $ f $ is pointwise only at $ (0,1]$ and we cannot speak about uniform convergence at $ [0,1]$. For $ a\in(0,1)$, the convergence is uniform at $ ...


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No, the convergence is not uniform. If a sequence of bounded functions converges uniformly to a function $f$, the $f$ is bounded too. But your function $f$ is unbounded.


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It is not clear to me how we can make this function continuous. I guess that we need to connect two points $(x,0)$ and $(1/2n, 2n)$ for $x < 1/2n$, and the slope of this line need to get steeper as $n$ goes to infinity. I am not sure how to write such function. Any help would be appreciated. One choice would be $$g^{(n)}(x) := \begin{cases} 2n & x\in[...


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Consider the pointwise linear map always vanishing except on $(1/2n, 1/n)$ such that $f(3/4n)=n$.


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By binomial theorem for any non-naturalindex, we have $$(1+z)^\nu=\sum_{k=0}^{\infty} {\nu \choose k} z^k.$$ Let $z=-x$ and $\nu=-n$, then $$(1-x)^{-n}=\sum_{k=0}^{\infty} {-n \choose k} (-x)^k$$ Now use the property that $$={-\nu \choose k}=(-1)^k {\nu+k-1 \choose k}$$ We get $$\sum_{k=0} {n+k-1 \choose k} x^k=(1-x)^{-n}$$ Replace $n$ by $n+1$ both sides, ...


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