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8

No. Look at $f(x) = \dfrac{1}{1 + x^2}$. Its Taylor series at $x = 0$ is just a geometric series with finite radius of convergence.


5

Note that at any choice of $x$ and for any integer $N$, there is an $n>N$ with $f_n(x)=1$. So, the numerical sequence $f_n(x)$ cannot converge to $0$. Note, however, that we can certainly select a subsequence of this sequence of functions that converges pointwise a.e.


3

Furthermore, even if the Taylor series converges, it does not necessarily converge to $f(x)$. Counter-example: Consider the function $f$ defined by: $$f(x)=\begin{cases}\mathrm e^{-\tfrac1{x^2}}&\text{if } x\ne 0\\0&\text{if } x= 0\end{cases}$$ On can prove by induction on the order of derivation that $f^{(n)}(x)=P_n\Bigl(\dfrac1x\Bigr)\mathrm ...


3

Divide numerator and denominator by $2^n$. It's not hard to show that $\frac{x^2}{2^n}\to 0$ (use L'Hôpital's rule, for example) to see $$\lim_{x\to\infty}\frac{2^{x+1}+(x+1)^2}{2^x+x^2}=\lim_{x\to\infty}\frac{2+(x+1)^2/2^x}{1+x^2/2^x}=2$$


3

In Gillman & Jerison's Rings of continuous functions I found the following note 8.21 N.B. A number of authors have fallen into the trap of assuming then every countable, closed, discrete subset of a completely regular space is $C^\ast$-embedded. We have just seen a counterexample: [...] It seems likely that one of these authors, or someone ...


3

Hint: For $n > x$ you can approximate $\sqrt{x + n} \le \sqrt{2}\sqrt{n}$


3

(1) follows from 'Portmanteau Theorem'. Denote probability measure induced by $X_n,X$ by $F_n$ and $F$ respectively. Note $\mathbb{Z}$ is closed in $\mathbb{R}$. As $X_n \overset{d}{\rightarrow} X$ we've $F(\mathbb{Z}) \geq \limsup F_n(\mathbb{Z})=1$ as $F_n(\mathbb{Z})=1$ $\forall n$ (2) You can say something strong namely $P(X_n=j) \rightarrow P(X)$ as ...


3

The Comparison Test says that, if $0 \le a_n \le b_n$, then If $\sum\limits_{n=1}^\infty b_n$ converges, then $\sum\limits_{n=1}^\infty a_n$ also converges. If $\sum\limits_{n=1}^\infty a_n$ diverges, then $\sum\limits_{n=1}^\infty b_n$ also diverges.


3

Hint: use the Borel-Cantelli lemma to show that $$P(X_n \ne 2 \text{ i.o.}) = 0.$$ (Independence is not needed,)


2

Take odd $f_n$s with $$f_n(x) = \begin{cases} nx, & \text{if $0\le x < 1/n$} \\ 2-nx, & \text{if $1/n\le x \le 2/n$.} \\ 0, & \text{if x > 2/n} \end{cases}$$ $f_n(x)\to 0$ pointwisely, since for any $x$ there is some $N_x$ that $x\in[-2/n,2/n]^c$ (namely $f_n(x)=0$) for any $n\ge N_x$. However the ...


2

Unlike integration, differentiation is a very unstable operation. It is very hard to make assumptions on $\{f_n\}_n$ so that $\{f'_n\}_n$ converges. For instance, let $f_n(x)= \frac{\sin (nx)}{n}$: $\{f_n\}_n$ converges to zero uniformly, but the derivatives $f'_n$ are oscillating. The only "elementary" theorem about differentiation of sequences of ...


2

hint: Show $x^{\frac{1}{x}} \to 1$ as $x \to \infty$. To this end, we have: $\dfrac{\ln x}{x} \to 0$ by L'hopitale rule hence the result follows.


2

$x^{\frac{2}{x}}=\exp\left(2\frac{\ln(x)}{x}\right)$, $\forall x>0$. Then, use the fact that $\frac{\ln(x)}{x}$ goes to $0$ as $x$ goes to $+\infty$.


2

First you can just focus on the series $\sum\frac{(-1)^n}{n}x^n$ since the series $\sum\frac{(-1)^n}{n}$ is convergent. Now this first series is an entire series and it's convergent on the interval $(-1,1]$.


2

Young's and Holder's inequalities make weak convergence easy to study, but the pointwise convergence depends on the behaviour of a maximal operator (the Carleson operator) for which it is not that easy to provide effective upper bounds. Carleson's greatest idea was probably to modify usual decomposition techniques in the Calderon-Zygmund theory in the ...


2

$$\lim_{x\to \infty}\frac{2\cdot (2^{x}+x^2)+(x+1)^2-2x^2}{2^x+x^2}$$ $$=2+\lim_{x\to \infty}\frac{(x+1)^2-2x^2}{2^x+x^2}$$ ($\frac{\infty}{\infty}$)form so using L-Hospital's rule twice: $$=2+\lim_{x\to \infty}\frac{-2}{2^x (ln2)^2+2}$$ $$=2$$


2

Notice, we have $$\lim_{x\to \infty}\frac{2^{x+1}+(x+1)^2}{2^x+x^2}$$ $$=\lim_{x\to \infty}\frac{2\cdot 2^{x}+x^2+2x+1}{2^x+x^2}$$ $$=\lim_{x\to \infty}\frac{(2^{x}+x^2)+(2^x+2x+1)}{2^x+x^2}$$ $$=1+\lim_{x\to \infty}\frac{2^x+2x+1}{2^x+x^2}$$ Using L-Hospital's rule 3 times: $$=1+\lim_{x\to \infty}\frac{2^x\ln 2+2}{2^x\ln 2+2x}$$ $$=1+\lim_{x\to ...


2

Draw a picture of the generic function $f_n$ in the typewriter sequence. It's a rectangle of height 1 over an interval of width $1/2^k$, with value zero elsewhere. As the sequence progresses, the rectangles slide across the unit interval, the way a typewriter moves across the page. At each 'carriage return' of the typewriter, a new row of rectangles starts, ...


2

Hint. A potential problem for convergence is near $x_1$. Since $x \mapsto V(x)$ is smooth, then as $x \to x_1^-$, by the Taylor expansion we have $$ V(x)=V(x_1)+(x-x_1)V'(x_1)+\mathcal{O}\left( x-x_1\right)^2 $$ or $$ V(x)=E_0-(x_1-x)V'(x_1)+\mathcal{O}\left( x_1-x\right)^2. \tag1 $$ Case 1. $V'(x_1)\neq0.$ Clearly, since $V(x)<E_0$ for any ...


1

When the sequence of event $(A_n)_n$ is increasing or decreasing this propriety is true.


1

If $p=\infty$ this is clear; fast convergence implies essentially uniform convergence. Suppose $p<\infty$. Fast convergence is much more than you need here. All you need is the weaker condition $$\sum||g_n-g||_p^p<\infty.$$That says $$\int\sum|g_n-g|^p<\infty.$$Hence the integrand is finite almost everywhere, which says (much more than) $g_n\to g$ ...


1

Two methods: Method I. Just to give a different approach, I'll write out the calculation via states. Our states: $S(0)$ is the starting state, no even numbers and no fives have been thrown. $S(1)$ is the state after one exactly one even number (but no fives) have been thrown. And we have Win and Loss states (where a Win here means a five is thrown ...


1

let y denote the expression given Then ln y= 1/2+(2/x)ln x. Then (2/x)ln x->0 as x->inf. Thus ln y ->1/2. Now take the exponential to get the answer.


1

As noted in a comment, you can settle the case for any $p \neq 1$ using comparison tests combined with the fact that the harmonic series diverges, and $\sum 1/n^p$ converges for $p > 1$. For $p = 1$, either use an integral comparison test, or if you note that $\ln n = (\log_2 n) / (\log_2 e)$ then you can work with the series $\sum_n 1/n \log_2 n$ and use ...


1

The formula for the Radius of Convergence of $$ \sum_{n=0}^\infty a_nx^n $$ is $$ R=\left(\limsup_{n\to\infty}\left|a_n\right|^{1/n}\right)^{-1} $$ This formula is derived using the Ratio Test. It is pretty easy to apply this to the series in the question. $$ \begin{align} \lim_{n\to\infty}\left(\frac1{(n+1)^2}\right)^{1/n} ...


1

An interesting aspect is to evaluate the series of the question. Consider \begin{align} S_{1}(x) &= \sum_{n=0}^{\infty} \frac{(-1)^{n} \, x^{n}}{(n+1)^{2}} \\ S_{2}(x) &= \sum_{n=0}^{\infty} (-1)^{n} \, (2^{n} + n^{2}) \, x^{n}. \end{align} The first series: \begin{align} \partial_{x} \left(x \, S_{1}(x) \right) &= \sum_{n \geq 0} \frac{(-1)^{n} ...



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