# Tag Info

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} ...

Only top voted, non community-wiki answers of a minimum length are eligible