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17

I'm sure there is a more elementary method than the heavy sledgehammer I used here. Since I only know this one, here is the sledgehammer. Let $\displaystyle c = \frac{1+i\sqrt{7}}{2}$, we are going to prove $\Re(c^k) \to \infty$ as $k \to \infty$. Consider the sequence $(c_k)_{k\in\mathbb{N}}$ where $c_k = 2\Re(c^k) = \left(\frac{1+i\sqrt{7}}{2}\right)^k + ... 16 Write$a = r e^{i\theta}$where$r = \sqrt{2}$and$\theta = \arctan(\sqrt{7})$. The question is whether for any$N$there exist infinitely many$n$such that$|2^{n/2} \cos(n \theta)| < N$. This can be viewed as a question about whether$2\theta/\pihas extremely good rational approximations. Namely, if \left| \dfrac{2\theta}{\pi} - \dfrac{m}{n} ... 10 Your suspicion about inner product is entirely correct. The trigonometic polynomials \{\sin(nx), \cos(mx)\} (possibly translated and scaled) are known to form an orthogonal system with respect to the scalar product given by \langle f, g\rangle:=\int fg, and if you choose the function space correctly (usually one uses a space called L^2) and use ... 8 It might help to break this up into smaller steps. \begin{align*} &V_0(y) = \sum_{n=1}^{\infty} C_n \sin \left(\frac{n \pi y}{a} \right) \\ \implies & V_0(y) \sin \left(\frac{n' \pi y}{a} \right) = \sum_{n=1}^{\infty} C_n \sin \left(\frac{n \pi y}{a} \right)\sin \left(\frac{n' \pi y}{a} \right) \\ \implies & \int_0^a V_0(y) \sin \left(\frac{n' ... 6 Assume \sum_{k=1}^{\infty} 1/n_k=\frac{p}{q} where p,q are strictly positive integer, Then, n_1n_2\cdots n_{k-1}\sum_{j=0}^{\infty}1/n_{k+j}\geq \frac{1}{q} $$for k\geq 2. Now let a_k=n_k^{\frac{1}{2^k}}, by hypothesis \lim_{n \to \infty} n_k^{1/2^k}=+\infty Therefore there exist r_1 such that a_j\leq a_{r_1} for j=1,2,\cdots,r_1-1 ... 6 No idea if this will help but could not resist posting these striking images. Try plotting the absolute value of the real part of a^n/2^{n/2} for 0\le n\le1000. Addendum. Even better, change the plot style to "point" so that you don't have lines joining successive points. 5 The function f(x)=a^x for a\geq 0 is uniquely determined by these three properties: \forall x,y \in\Bbb R\left[f(x+y)=f(x)f(y)\right] f is continuous at at least one point f(1)=a What MartÃ­n-Blas is trying to convey is that, once you agree that the fundamental property that an 'exponential-like' operation E:\Bbb R\rightarrow \Bbb R has is ... 5 There is no need for complicated Cauchy products. We will simply use \Big(f^2(x)\Big)'=2f(x)f'(x).$$\sin(x)=\sum_0^\infty(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}\iff\sin'(x)=\sum_0^\infty(-1)^{n}\frac{x^{2n}}{(2n)!}=\cos(x)\cos(x)=1+\sum_1^\infty(-1)^{n}\frac{x^{2n}}{(2n)!}\iff\cos'(x)=\sum_1^\infty(-1)^{n}\frac{x^{2n-1}}{(2n-1)!}=-\sin(x)$$Now, ... 5 From the 20th entry onward, we reach the following stable pattern, wherein the first digit follows a cycle of length 8:$$ 2 \overbrace{0\cdots 0}^n,\\ 3 \overbrace{1\cdots 1}^n,\\ 4 \overbrace{0\cdots 0}^n,\\ \vdots\\ 9 \overbrace{1\cdots 1}^n,\\ 2 \overbrace{0\cdots 0}^{n+1} $$It clearly follows that$$ \lim_{k \to \infty}\frac{a_{k+8}}{a_k} = 10 $$... 5 Not necessarily. Consider the series$$\frac{1}{\sqrt[3]{2}}-\frac{1}{2\sqrt[3]{2}}-\frac{1}{2\sqrt[3]{2}}+\frac{1}{\sqrt[3]{3}}-\frac{1}{2\sqrt[3]{3}}-\frac{1}{2\sqrt[3]{3}}+\frac{1}{\sqrt[3]{4}}-\frac{1}{2\sqrt[3]{4}}-\frac{1}{2\sqrt[3]{4}}+\frac{1}{\sqrt[3]{5}}-\frac{1}{2\sqrt[3]{5}}-\frac{1}{2\sqrt[3]{5}}+\cdots.$$It is clear that this converges. ... 5 Using Bernoulli's Inequality, we see that$$ e^{-k}\le\left(1-\frac{k}{n}\right)^{n-k}\le e^{-k(1-k/n)}\tag{1} $$Thus,$$ \begin{align} \lim_{n\to\infty}\frac1{e^n}\sum_{k=1}^n\binom{n}{k}\left(\frac{ck}{n}\right)^k &=\lim_{n\to\infty}\frac1{e^n}\sum_{k=0}^n\binom{n}{k}\left(\frac{c(n-k)}{n}\right)^{n-k}\\ ... 4 Use integration by parts as follows$$\langle f, e_n \rangle=\frac{1}{\sqrt{2\pi}}\int\limits_{-\pi}^{\pi} x e^{inx}\, dx \\= \frac{1}{\sqrt{2\pi}}\left[\left.x\cdot \frac{e^{inx}}{in}\right|_{-\pi}^{\pi}-\int\limits_{-\pi}^{\pi} \frac{e^{inx}}{in}\cdot 1\, dx\right]$$ Added The integral will be $$c_n=\frac{1}{\sqrt{2\pi}}\cdot \frac{2\pi\cos n\pi}{in} \\ ... 4$$ S=\sum \limits_{i = 1}^n i(i+1)(i+2) \dots (i + k) = \sum_{i=1}^{n}\frac{(i+k)!}{(i-1)!}  \frac{S}{(k+1)!}=\sum_{i=1}^{n}\frac{(i+k)!}{(i-1)!(k+1)!}=\sum_{i=0}^{n-1}\binom{i+k+1}{i}  \frac{S}{(k+1)!}=\binom{k+1}{0} + \binom{k+2}{1} + \dots + \binom{n+k}{n-1} \\ =\binom{k+2}{0} + \binom{k+2}{1} + \dots + \binom{n+k}{n-1} \\ =\binom{k+3}{1} + ... 4 From what I understand, your question comes down to why a set being ordered does not imply it being countable. The slightly subtle notion here is the difference between cardinal and ordinal numbers. In general, there are two standard ways to compare the sizes of two sets. You can construct a bijection (as you've probably seen), or you can construct an ... 3 The series\displaystyle\sum_{n\ge0} a_n$is convergent if and only if the sequence of partial sum$(S_n)_n =\left(\displaystyle\sum_{k=0}^n a_k\right)_nis convergent and in this case $$\lim_{n\to\infty}S_{n}-S_{n-1}=\lim_{n\to\infty}a_n=0$$ and simply we get the desired result by contrapositive. 3 The statement is not true in the sense that it is provable. The statement is true in the sense that as a definition, it is consistent (not contradictory, not paradoxical). Or more exactly, it is a special case of an generalized definition of summations. You are probably familiar with the definition $$\sum_{n=1}^\infty f(n) = \lim_{k\rightarrow \infty} ... 3 This should get you started (or more).$$ \begin{align} \sum_{k=1}^n\log\left(\frac{k(k+2)}{(k+1)^2}\right) &=\sum_{k=1}^n\log\left(\frac{k}{k+1}\right)-\sum_{k=1}^n\log\left(\frac{k+1}{k+2}\right)\\ &=\sum_{k=1}^n\log\left(\frac{k}{k+1}\right)-\sum_{k=2}^{n+1}\log\left(\frac{k}{k+1}\right)\\ ... 3 Notice that forx=1$the series$\sum_n \frac1{n+3}$is divergent then the radius of convergence$R\le1$but for$x=-1$the series$\sum_n \frac{(-1)^n}{n+3}$is convergent by Leibniz theorem so$R\ge1$. Conclude. For the sum we have $$\sum_{n=0}^\infty \frac{x^n}{n+3}=\sum_{n=3}^\infty \frac{x^{n-3}}{n}=\frac1{x^3}\sum_{n=3}^\infty ... 3 Here are a couple of ideas - though I think the problem may be harder than it looks. Let a=r(\cos \theta+i\sin \theta) so that a^n=r^n (\cos n\theta+i\sin n\theta) and u_n=r^n \cos n\theta. You will find that r=\sqrt 2 and \theta = \arctan \sqrt 7. This shows that the general growth is like (\sqrt 2)^n but you have to bound the trigonometric ... 3 Your first instinct should be, "[a,b] is compact so of course sequences of functions behave nicely." ;) A general theme is that things are nice on compact spaces, so we'll want to use the compactness of [a,b] to get this result. You have a small interval about each c\in[a,b] in which f_n converges uniformly. This gives you an open cover of [a,b], ... 3 The series is given by$$1 + \sum_{k=1}^\infty (-1)^k \left(\frac1{2k} + \frac1{2k+1}\right)$$Assuming convergence we split it up into$$1 + \sum_{k=1}^\infty (-1)^k \frac1{2k} + \sum_{k=1}^\infty (-1)^k \frac1{2k+1}$$Both parts converge due to leibniz and thus their sum converges as well. Note that pulling apart series is not allowed in general, but ... 3 Yes there is. Ever wonder why this is called quadratic sequence? Quadratic refers to squares right? This is just constant difference of difference. So where's the connection? Well as it turns out, all terms of a quadratic sequence are expressible by a quadratic polynomial. What do I mean? Consider this$$ t_n = n+n^2 $$Subsituiting n=1,2,3,\cdots ... 3 As$$0 \le \frac{x_n}{1+x_n}\le x_n,\\ \sum {x_n}<\infty\Rightarrow \sum \frac{x_n}{1+x_n}<\infty $$Now suppose that \sum \frac{x_n}{1+x_n}<\infty . \frac{x_n}{1+x_n}\to 0, and so x_n\to 0, so there is a N such as n>N\Rightarrow x_n<1 and then$$ \frac{x_n}{1+x_n}>\frac {x_n}2 $$so$$ \sum \frac{x_n}{1+x_n}<\infty ... 3 Hint: These seem to be exercises related to Riemann Sums. For example,$1.$$$\sum_{k=1}^{n-1}\sqrt{k/n}\frac1n$$ is a Riemann Sum for $$\int_0^1\sqrt{x}\,\mathrm{d}x$$$2.$$$\sum_{k=1}^n\log\left(1+\frac {k^2}{n^2}\right)\frac1n$$ is a Riemann Sum for $$\int_0^1\log(1+x^2)\,\mathrm{d}x$$ 3$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} ...

3

I was also working with the inequality $$\left(1 - \frac{k}{n}\right)^{n-k} \leq \exp \left( -k + \tfrac{k^{2}}{n} \right) = \exp\left\{ -n \cdot \tfrac{k}{n} \left( 1 - \tfrac{k}{n} \right) \right\}.$$ Now note that $q(x) = x(1-x)$ satisfies $q(1-x) = q(x)$ and $q(x) \geq \frac{1}{2}x$ on $[0, \frac{1}{2}]$. Then we get $$0 \leq k \leq \tfrac{1}{2}n ... 3 To get an idea of the asymptotics, try using Stirling's approximation on the binomial coefficients:$$\ln {n \choose k} \approx n \ln n - k \ln k - (n - k) \ln(n - k).$$Combining this with the other term in the summation,$$\ln \left(\frac{c k}{n}\right)^k = k \ln c + k \ln k - k \ln n,$$we get$$\ln \left[{n \choose k} \left(\frac{c ...

3

Clearly $$\frac{1}{(k-1)k(k+1)}=\frac{1}{2}\left(\frac{1}{k(k-1)}-\frac{1}{k(k+1)}\right).$$ Thus $$4\sum_{k=1}^{n-1} \frac{1}{k(k+1)(k+2)}=2\sum_{k=1}^{n-1}\left(\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)}\right)=\frac{2}{1\cdot 2}-\frac{2}{n(n+1)}.$$ Thus $$\lim_{n\to\infty}4\sum_{k=1}^{n-1} \frac{1}{k(k+1)(k+2)}=1.$$

2

Expand by partial fractions: $$\frac{1}{(n - 1)n(n + 1)} = -\frac 1 n + \frac 1 2 \frac 1 {n - 1} + \frac 1 2 \frac 1 {n + 1}$$ Now notice that a lot of the terms will cancel, and the series telescopes. This is similar to, and motivated by the fact that $$\sum_{n = 1}^{\infty} \frac 1 {n(n + 1)} = \sum_{n = 1}^{\infty} \frac 1 n - \frac 1 {n + 1}$$ ...

2

Suppose for the sake of contradiction that there is no real positive $n$ such that $$(n+1) \left (1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\cdots\dfrac{1}{n+1} \right ) \geq 5280n.$$ Then we would have, for all $n$, $$\left (1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\cdots\dfrac{1}{n+1} \right ) < 5280\frac{n}{n+1}<5280,$$ implying that the infinite ...

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