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12

We have that $$\left(1+\frac{1}{n}\right)^n=\mathrm{e}^{n\log(1+\frac{1}{n})}=\mathrm{e}^{1-\frac{1}{2n}+{\mathcal O}(n^{-2})}=\mathrm{e}\left(1-\frac{1}{2n}+{\mathcal O}\Big(\frac{1}{n^2}\Big)\right),$$ since $$\log \Big(1+\frac{1}{n}\Big)=\frac{1}{n}-\frac{1}{2n^2}+{\mathcal O}\Big(\frac{1}{n^3}\Big),$$ and hence $$... 3 We know that we can make the term |f_n(x_n) - f(x_n)| small since \{f_n\} converges uniformly. We know that we can make |f(x_n)-f(x_0)| small since f is continuous, and x_n \to x_0. Using this information, we can make |f_n(x_n)-f(x_0)| small, which is what we want to do to show that \lim f_n(x_n) = f(x_0). Edit: To be more specific, let ... 2 A sequence X_1,X_2,\ldots of random variables is said to converge in distribution, or converge weakly, or converge in law to a random variable X if$$ \lim_{n\to\infty}F_n(x)=F(x) $$for every number x\in\mathbb R at which F is continuous, where F_n(x)=\mathbb P(X_n\le x) and F(x)=\mathbb P(X\le x). Thus, we need to show that F(x) is ... 2 Note that T_n=uX_n - \frac12nu^2\sigma^2 defines a random walk (T_n) whose steps have mean E(uY_1- \frac12u^2\sigma^2)=- \frac12u^2\sigma^2\lt0 if u\ne0 hence, for every u\ne0, T_n\to-\infty almost surely, Z^u_\infty=0 almost surely and Z^u_n\ne E(Z^u_\infty\mid\mathcal F_n). The case u=0 is direct. 2 Let w=\dfrac{\partial v}{\partial x}. Since v\in W^{1,\infty}, we have w\in L^\infty. The space L^\infty is dual to L^1. Thus, pairing a fixed L^\infty function with a convergent sequence of L^1 functions produces a convergent sequence of numbers. This is why$$\int b(u_k) w\to \int b(u)w$$For the other integral, note that a(u_k)w ... 1 Part 1 I don't think your answer to this is correct. You need to justify why we can be sure that g has a power series that converges if |z|<1, and you haven't done that. You have correctly shown that g is a product of two power series:$$g(z)=\sum_{k=0}^{\infty}a_kz^k *\sum_{n=0}^{\infty}z^n$$We also know that f(z)=\sum_{k=0}^{\infty}a_kz^k has ... 1 For (a_n) the pointwise limit is a(x)=x^2+3 and we have$$||a_n-a||_1=\int_0^1 \left|\frac{-x^2}{n+1}+\frac{2}nx\right|dx\le\frac{1}{3(n+1)}+\frac{1}{n}\xrightarrow{n\to\infty}0$$hence$$a_n\xrightarrow{||.||_1}a$$by the same method prove that$$b_n\xrightarrow{||.||_2}(0,1,1)$$1 a_n = \dfrac{a_{n-1}+a_{n-2}}{2}. One overkill way of finding the limit, is getting the a_n in terms of n via a generating function. Write F(x) = \sum_{i=0}^{\infty}a_i x^i. Thus,$$ \frac{F(x)-a_0}{x}=\sum_{i=0}^{\infty}a_{i+1}x^{i}  \frac{F(x)-a_0-a_1}{x^2} = \sum_{i=0}^{\infty}a_{i+2}x^{i} $$So,$$ \frac{F(x)-a_0-a_1}{x^2} = ...

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Start with the study of $$1-\frac nx\sin\frac xn$$on $(0,2)$. It is $F(\frac xn)$ where $F(x) = 1-\frac 1x\sin x$. As $\lim_0 F = 0$, for each $\epsilon>0$ one can find $r>0$ such as $$|x|<r\Rightarrow |F(x)|<\epsilon$$ then taking $N>\frac 2r$,$$n\ge N \ \ \&\ \ 0<x<2\Rightarrow \frac xn<\frac 2N<r$$ and $$\left|1-\frac ... 1 This is not a full answer, but an approach which may help in developing a complete answer. Let's first replace x by the usual symbol q and I will just stick to the real values of q. Clearly we have$$\begin{aligned}Q(-q) &= \prod_{k = 1}^{\infty}(1 + (-q)^{k})\\ &= \prod_{k = 1}^{\infty}(1 + q^{2k})(1 - q^{2k - 1})\\ &= \prod_{k = ...

1

Using Cauchy condensation, if $\displaystyle \sum_{n = 2}^{\infty} \frac{2^n}{2^n \log 2^n}$ converges or diverges, then the same must be true of my desired series. This series is equal to $\frac{1}{\log 2} \displaystyle \sum_{n = 2}^{\infty} \frac{1}{n}$, the harmonic series, thus it diverges, and so does my desired series. The comparison aspect of this ...

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