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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 ...


1

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 ...


0

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 $f(x_n)\to 1\ne f(0)$


1

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 $\rho,\theta$ are 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. Let $f$ 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 $V(x)<E_0$ for 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 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 $$\lim_{x\to\infty}\frac{2^{x+1}+(x+1)^2}{2^x+x^2}=\lim_{x\to\infty}\frac{2+(x+1)^2/2^x}{1+x^2/2^x}=2$$


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 ...


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.


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 ...


4

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


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 ...


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 ...


0

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...


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

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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