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0

You can show that $x\mapsto\frac{2^x-1}{x}$ is an increasing function, which would imply that the left and right-hand limits exist at $x=0$. To do that, it suffices to show that $x\mapsto 2^x$ is convex. Since the mapping is continuous, it suffices to show midpoint convexity, i.e. $$2^{\frac{x_1+x_2}{2}}\le \frac{2^{x_1}+2^{x_2}}{2}$$ which is true by ...

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If you omitted condition (b), then nothing in the hypothesis tells you what $f$ is. Remember that a theorem should be correct no matter what particular values you give its variables; as long as the hypotheses are true, the conclusion must be true also. Now suppose you chose some reasonable functions as your $f_n$'s, satisfying hypothesis (a), so they ...

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For part b, all we need is a noncontinuous function, since continuity is what grants us that property. Here's a simple one: Define $f(x)=1$ if $x\ne 0$ and $f(0)=0$. Then look at $x_n=\frac 1 n$. Clearly $x_n\to 0$, but $f(x_n)=1$ for all $x_n$, hence $f(x_n)\to 1\ne f(0)$

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Hint: A subsequence of a convergent sequence is convergent. Conversely, setting $z_{2n - 1} = x_n$ and $z_{2n} = y_n$. As $\lim x_n = \lim y_n = a$, for any given $\varepsilon > 0$ there exists $n_1, n_2 \in \mathbb N$ such that $$n > n_1 \implies |x_n - a| < \varepsilon \\ n > n_2 \implies |x_n - a| < \varepsilon$$ Consider $n_0 = ... 2 Hint: For$T(n)=\frac{n(n-1)}{2}$we have$\frac{1}{T(n)}=2\left(\frac{1}{n-1}-\frac{1}{n}\right)$. And please correct the statement "the two series are equal." You mean to say the two series' sums are equal. 1 We have the following: $$2^{-i} = \Bigg(\frac{1}{2}\Bigg)^i$$ And thus: $$\sum_{i=0}^\infty 2^{-i} =\sum_{i=0}^\infty \Bigg(\frac{1}{2}\Bigg)^i$$ Now you can use your properties for infinite sums that you have, if your base is$<1$Infact, what you have here, is called geometric series by mathematicans. Look it up on wikipedia. 0 Based on the comments to the question: Let$z=a+bi\in\mathbb{C}$. We want to determine where in the complex plane the series $$S(z)=\sum_{n=0}^\infty z^n$$ converges. Using the root test: $$\lim_{n\rightarrow\infty}\sqrt[n]{|z^n|}=|z|$$ we see that the sum absolutely converges if$|z|=|a+ib|=\sqrt{a^2+b^2}<1$, and the sum diverges if$|z|>1$. Thus ... 1 HINT: write$z=a+ib$as$z=\rho e^{i\theta}$where$\rho,\thetaare polar coordinates. So $$\lim_{n\longrightarrow \infty}(a+ib)^n=\lim_{n\longrightarrow \infty}\rho^ne^{in\theta}$$ 4 Hint: write $$\lim\limits_{x\to 0}\frac{2^x-1}{x}=\lim\limits_{x\to 0}\frac{2^x-2^0}{x-0}.$$ This should hopefully look familiar to you (if not: think of the definition of the derivative). 3 We can appeal to the sum of a geometric series as follows: \begin{align} \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2^{2n-1}}&=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2^{2n+1}}\\\\ &=\frac12\sum_{n=0}^{\infty}\left(\frac{-1}{4}\right)^n\\\\ &=\frac12\frac{1}{1+\frac14}\\\\ &=\frac25 \end{align} 1 It was already answered to your question (and brilliantly). I use this space only for some variation on the theme. Your post reminds me of a method of proving the existence of an antiderivate of a continuous function. Letf$be continuous on$[a,b]$. Now it is well known there exists a sequence of piecewise constant functions that converges uniformly to ... 1 Since$\sin \frac{\pi}{n} \sim \frac{\pi}{n}$as$n \to \infty$, and since$\sum_{n \geq 1}\frac{\pi}{n}$diverges, by the limit comparison test we see that$\sum_{n\geq 1}\sin \frac{\pi}{n}$diverges. 7 For$x\le\frac\pi2$, concavity implies$\frac2\pi x\le\sin(x)\le x. Therefore, \begin{align} \sum_{n=1}^\infty\sin\left(\frac\pi n\right) &=\sum_{n=2}^\infty\sin\left(\frac\pi n\right)\\ &\ge\frac2\pi\sum_{n=2}^\infty\frac\pi n\\ &=2\sum_{n=2}^\infty\frac1n \end{align} which diverges. Furthermore, $$\lim_{n\to\infty}\sin\left(\frac\pi ... 2 sin(\frac{\pi}{n}) is asymptotically equivalent to \frac{\pi}{n} so it behaves like the harmonic series which is divergent. 2 Hint : Take , v_n=\frac{1}{n}. Then use comparison test. 0 The latest publication I know of on this subject is http://arxiv.org/abs/0806.4410 This article has references to many older articles on the subject. This article also has a Mathematica package, kempnerSums.m, that can calculate the sum of 1/n where n has no occurrence of, say, 32947902384769234. In fact, using the function kSum[ ] in the package, this ... 2 Your hypothesis is never satisfied except in the trivial case when ||\mu_n-\mu||\le ce^{-n}. (Which actually means the answer to your question is yes, but it's not the sort of yes I suspect you wanted.) Theorem Suppose K is a compact Hausdorff space, a_n>0, and for every f\in C(K) we have |\mu_n(f)-\mu(f)|\le c_f a_n. Then ||\mu_n-\mu||\le ... 0 Just to expand on Alex R.'s observation: You can determine that closed form by determining the sequence's generating function. Suppose F(x)=\sum\limits_{n\ge0}f(n)x^n.$$\begin{align*} \sum_{n\ge1}f(n)x^n&=\frac{1}{2}\sum_{n\ge1}f(n-1)x^n+\frac{1}{16}\sum_{n\ge1}x^n\\[1ex] ... 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 ... 0 We say Ignatz wins if he rolls two even numbers before a$5$. What is the probability he wins in turn$n+2$? It is equal to the probability he rolls an even number in turn$n+2$and exactly one even number before turn$n+2$. There are$2^{n}$ways to through the odd numbers ,$n+1$ways to select the turn in which the first even number is thrown and$3$... 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, ... 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. 0 Basically the same as David Ullrich's proof, but maybe a little shorter: Suppose$f\in C^1 [a,b].$Let$\epsilon>0.$By the uniform continuity of$f'$on$[a,b],$there exists$\delta > 0$such that$|y-x|<\delta \implies |f'(y)-f'(x)| < \epsilon.$Let$P = \{x_0,\dots x_n\}$be a partition of$[a,b]$of mesh size less than$\delta.$Define ... 0 You ask about$\phi_n\to\phi$, but I gather what you're actually concerned with is$\phi_n'\to\phi'$. This happens at every point of$N^c$, just from the definition of the derivative, more or less. If$x\in N^c$then for every$n$there exists$i$such that $$x\in(i/n,(i+1)/n)=(a_n, b_n).$$ Also $$\phi_n'(x)=\frac{\phi(b_n)-\phi(a_n)}{b_n-a_n}.$$ Hence ... 0 For the radius of convergence, we have $$R=\lim_{n\to\infty}\left| \frac{(-1)^{n+1}x^{5n+7}}{5n+7}\frac{5n+2}{(-1)^nx^{5n+2}} \right| \\ =\lim_{n\to\infty}\left| \frac{(-1)x^5(5n+2)}{5n+7} \right| \\ =|x|^5\lim_{n\to\infty}\frac{5n+2}{5n+7} \\ =|x|^5\lim_{n\to\infty}\frac{5+\frac{2}{n}}{5+\frac{7}{n}}=|x|^5.$$ We will have convergence only if$|x|^5<1.$... 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 ... 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)$g_n\to g$... 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 ... 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. 0 Hint. As Carl Heckman noticed, you have probably mixed up sequences and series. For series$\sum a_n$and$\sum b_n$, such that $$0\leq a_n\leq b_n,$$ the following holds true: If the series$\sum a_n$diverges then the series$\sum b_n$diverges If the series$\sum b_n$converges then the series$\sum a_n$converges 0 The result you're trying to prove is false. Let$a_n = (-1)^n + 2$, and$b_n = 5$. Clearly$b_n \to 5$, but$a_n$oscillates between$3$and$1$without converging. You've basically mimicked that example, but with some functions (strictly, you should evaluate your functions at each natural number in order to get a true sequence). There is a related result, ... 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_0for any ... 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} ... 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\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 ... 3 Divide numerator and denominator by2^n$. It's not hard to show that$\frac{x^2}{2^n}\to 0$(use L'Hôpital's rule, for example) to see $$\lim_{x\to\infty}\frac{2^{x+1}+(x+1)^2}{2^x+x^2}=\lim_{x\to\infty}\frac{2+(x+1)^2/2^x}{1+x^2/2^x}=2$$ 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. 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$+\infty$. 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 ...

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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.

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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 ...

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

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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 ...

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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 ...

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Take the sequence: $$f_n(x) = \begin{cases} n\qquad \mbox{if }\; \frac{1}{n+1}\leq x< \frac{1}{n} \\ 0 \qquad \mbox{otherwise}\end{cases}$$ We have $\int_{0}^1 f_n(x) dx = \frac{1}{n+1}$. So $\int_0^1 f_n(x) dx \to 0$. What happens to $g(x)=\max\{f_n(x): n\in \mathbb{N}\}$? Now you need to smoothen $f_n$ a little bit, to make them continuous...

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(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 ...

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

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